Chapter 11. Answers of Circles

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Given circle is $x^{2} + y^{2} - \frac{8}{3} x - \frac{10}{3} y + 1 = 0$ . We know that if the equation is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , then the center is $(- g, - f)$ and radius is $\sqrt{g^{2} + f^{2} - c}$ .

Thus, center of circle is $(\frac{4}{3}, \frac{5}{3})$ and radius is $\sqrt{\frac{16}{9} + \frac{25}{9} - 1} = \frac{4}{3} \sqrt{2}$ .

Radius of $x^{2} + y^{2} = 1$ is $1$ . Radius of $x^{2} + y^{2} - 2 x - 6 y = 6$ is $\sqrt{1^{2} + 3^{2} + 6} = 4$ . Radius of $x^{2} + y^{2} - 4 x - 12 y = 9$ is $\sqrt{2^{2} + 6^{2} + 9} = 7$ .

Thus, radii of the given circles are in A.P. with a common difference of $3$ .

The center will be $C (- g, - f)$ and the radius will be $\sqrt{g^{2} + f^{2} - c}$

From $△ Q L C, Q L = C Q \sin 60^{\circ} = \frac{\sqrt{3}}{2} \sqrt{g^{2} + f^{2} - c}$

Figure 11.1.

$Q R = 2 Q L = \sqrt{3} \sqrt{g^{2} + f^{2} - c}$

$Δ = \frac{\sqrt{3}}{4} 3. (g^{2} + f^{2} - c) = \frac{3 \sqrt{3}}{4} (g^{2} + f^{2} - c)$ .

The center of the circle is $(3, - 4)$ and radius is $\sqrt{3^{2} + (- 4)^{2} + 25} = 7$ .

Given circle is $x^{2} + y^{2} + \frac{4}{5} x - \frac{8}{5} y = \frac{16}{5}$ . Center is $(- \frac{2}{5}, \frac{4}{5})$ and radius is $\sqrt{\frac{(- 2)^{2} + 4^{2} + 80}{25}} = 2$ .

Center is $(3, 1)$ and radius is $\sqrt{3^{2} + 1 + 6} = 4$ .

Given circle is $x^{2} + y^{2} + 2 x \cos θ + 2 y \sin θ - 8 = 0$ . Center is $(- \cos t h e t a, - \sin θ)$ and radius is $\sqrt{\cos^{2} θ + \sin^{2} θ + 8} = 3$ .

Centers of the given circles are $(0, 0), (- 3, 1)$ and $(6, - 2)$ . Area of the triangle whose vertices are these three points is given by

$Δ = \frac{1}{2} | \begin{matrix} 0 & 0 & 1 \\ - 3 & 1 & 1 \\ 6 & - 2 & 1 \end{matrix} | = 0$ .

Hence, the centers of the given circles are collinear.

Radii are $1, \sqrt{1 + 3^{2} + 6} = 4$ and $\sqrt{2^{2} + 6^{2} + 9} = 7$ . Common difference is $3$ , and hence, the given circles have radii in A.P.

Radii are $2, \sqrt{1^{2} + 3^{2} - \frac{15}{4}} = \frac{5}{2}$ and $\sqrt{2^{2} + 5} = 3$ . Common difference is $\frac{1}{2}$ , and hence, the given circles have radii in A.P.

The line $3 x = 4 y + 15$ gives $y = \frac{3 x - 15}{4}$ .

Substituting into the circle equation $x^{2} + y^{2} = 9 + 4 r^{2}$ gives $x^{2} + {(\frac{3 x - 15}{4})}^{2} = 9 + 4 r^{2}$ .

Multiplying through by $16$ gives $16 x^{2} + (3 x - 15)^{2} = 16 (9 + 4 r^{2})$ .

Expanding gives $16 x^{2} + 9 x^{2} - 90 x + 225 = 144 + 64 r^{2}$ .

Simplifying gives $25 x^{2} - 90 x + 81 - 64 r^{2} = 0$ .

Solving for $x$ gives $x = \frac{90 \pm \sqrt{8100 - 100 (81 - 64 r^{2})}}{50}$ .

This simplifies to $x = \frac{90 \pm 80 r}{50} = \frac{9 \pm 8 r}{5}$ .

Thus, the two points of intersection correspond to $x = \frac{9 - 8 r}{5}$ and $x = \frac{9 + 8 r}{5}$ .

The length of the intercept cut by each circle on the line is $\frac{(9 + 8 r) - (9 - 8 r)}{5} = \frac{16 r}{5}$ .

For successive values of $r$ , the difference of intercepts is constant since $\frac{16 \times 2}{5} - \frac{16 \times 1}{5} = \frac{16}{5}$ and $\frac{16 \times 3}{5} - \frac{16 \times 2}{5} = \frac{16}{5}$ .

Therefore, the circles cut off equal intercepts on the given line.

Since point $(4, 6)$ lies on the circle, therefore, radius $= \sqrt{(1 - 4)^{2} + (2 - 6)^{2}} = 5$ .

Thus, equation of the circle would be $(x - 1)^{2} + (y - 2)^{2} = 5^{2} \Rightarrow x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ .

Solving the equations of the diameters we obtain the center as $(8, - 2)$ .

Thus, equation of the circle is given by $(x - 8)^{2} + (y + 2)^{2} = 10^{2} \Rightarrow x^{2} + y^{2} - 16 x + 4 y - 32 = 0$ .

Since the line touches the circle, therefore, the perpendicular distance from center will give radius.

Radius is $\frac{3.5 + 4.12 - 1}{\sqrt{5^{2} + 12^{2}}} = \frac{62}{13}$ .

Thus, equation of the circle is $(x - 3)^{2} + (y - 4)^{2} = {(\frac{62}{13})}^{2} \Rightarrow x^{2} + y^{2} - 6 x - 8 y + \frac{381}{169} = 0$ .

Solving the two equation we have $x = \frac{x + 1}{3 c + 2}$ . Now as $c \to 1, x = \frac{2}{5} \Rightarrow y = - \frac{1}{25}$ .

The circle passes through $(2, 0)$ , so the radius is $\sqrt{{(2 - \frac{2}{5})}^{2} + {(\frac{1}{25})}^{2}}$

Thus, equation of the circle is ${(x - \frac{2}{5})}^{2} + {(y + \frac{1}{2} 5)}^{2} = \frac{64}{25} + \frac{1}{625} \Rightarrow 25 (x^{2} + y^{2}) - 20 x + 2 y - 60 = 0$ .

Let the center be $(α, β)$ . Since it lies on the line $y = x - 1 ∴ β = α - 1$

Equation of the circle would be $(x - α)^{2} + (y - β)^{2} = 3^{2}$ . Putting the value of $β$ from the above equation, we have

$(x - α)^{2} + (y - α + 1)^{2} = 9$

Since it passes through $(7, 3)$ , therefore, $(7 - α)^{2} + (4 - α)^{2} = 9 \Rightarrow a l p h a^{2} - 11 α + 28 = 0$

$α = 4, 7 \Rightarrow β = 3, 6$

So the center is $(4, 3)$ or $(7, 6)$ .

Thus equation of circle is $(x - 4)^{2} + (y - 3)^{2} = 3^{2}$ or $(x - 7)^{2} + (y - 6)^{2} = 3^{2}$ i.e. $x^{2} + y^{2} - 8 x - 6 y + 16 = 0$ or $x^{2} + y^{2} - 14 x - 12 y + 76 = 0$ .

Since the circle touches the axes the center will lie on $x = y$ or $y = - x$ .

Case I: When the center lies on the line $y = x$ .

Solving the two equations we get $x = - 3, y = - 3$

The radius of the circle would be $3$ . Therefore, the equation of the circle is

$(x + 3)^{2} + (y + 3)^{2} = 3^{2} \Rightarrow x^{2} + y^{2} + 6 x + 6 y + 9 = 0$ .

Case II: When the center lies on the line $y = - x$ .

Solving the two equations we get $x = 1, y = - 1$ .

The radius of the circle will be $1$ . Hence, equation of the curcle is $(x - 1)^{2} + (y + 1)^{2} = 1 \Rightarrow x^{2} + y^{2} - 2 x + 2 y + 1 = 0$ .

Radius of the first circle is $2$ so its center would be $(2, 2)$ as it touches both the coordinate axes.

Figure 11.2.

Distance between the centers is $\sqrt{(2 - 6)^{2} + (2 - 5^{2})} = 5$ . Hence, radius of the second circle is $5 - 2 = 3$

Hence, equation of the circle is $(x - 6)^{2} + (y - 5)^{2} = 3^{2} \Rightarrow x^{2} - 12 x + 36 + y^{2} - 10 y + 25 = 9$

$\Rightarrow x^{2} + y^{2} - 12 x - 10 y + 52 = 0$ .

Half of intercept(chord) is $3$ . Perpendicualr distance from center to $2 x - 5 y + 18 = 0$ is

$\frac{2.3 - 5. (- 1) + 18}{\sqrt{2^{2} + 5^{2}}} = \sqrt{29}$

Thus, radius of the circle is $\sqrt{{(\sqrt{29})}^{2} + 3^{2}} = \sqrt{38}$ .

Since the circle touches $y$ -axis at $(0, 3)$ , therefore, the two circles will be in first two quadrants. Let $C$ and $D$ be the centers of the circles in the first and second quadrant respectively. Let $A B$ be the intercept on $x$ -axis i.e. $A B = 8$ . Let $C L$ be the perpendicular from $C$ on $x$ -axis.

$∴ A L = 4. O P = 3 ∴ C L = 3$

$A C = \sqrt{4^{2} + 3^{2}} = 5 ∴ C = (5, 3)$ and $D = (- 5, 3)$

Hence, equations of circles are $(x - 5)^{2} + (y - 3)^{2} = 5^{2}$ and $(x + 5)^{2} + (y - 3)^{2} = 5^{2}$ i.e

$x^{2} + y^{2} - 10 x - 6 y + 9 = 0$ and $x^{2} + y^{2} + 10 x - 6 y + 9 = 0$ .

$x^{2} + 2 x y + 3 x + 6 y = 0 \Rightarrow (x + 3) (x + 2 y) = 0 \Rightarrow x + 3 = 0$ and $x + 2 y = 0$

Since these are normals, therefore, their point of intersection will be the center of the circle. Solving gives us $(- 3, \frac{3}{2})$ as the center of the circle.

Center of the given circle is $(2, \frac{3}{2})$ and its radius is $\sqrt{2^{2} + \frac{9}{4}} = \frac{5}{2}$

Distance between centers is $5$ , which is greater than radius of the given circle.

Since the circle is just large enough to contain the given circle the radius of the required circle is distance between the centers plus radius of the given circle i.e. $5 + \frac{5}{2} = \frac{15}{2}$

Thus, the equation is $(x + 3)^{2} + {(y - \frac{3}{2})}^{2} = {(\frac{15}{2})}^{2} \Rightarrow x^{2} + y^{2} + 6 x - 3 y - 45 = 0$ .

Let $C$ be the center of the circle in original position adn $D$ be the center of its new position. Then $C = (5, 5)$ and $D = (5 + 10 π, 5)$ . Radius is $5$ .

Figure 11.3.

Thus, equation of the circle in new position is $(x - 5 - 10 π)^{2} + (y - 5)^{2} = 5^{2}$

$=> x^{2} + y^{2} - 10 (2 π + 1) x - 10 y + 100 π^{2} + 100 π + 25 = 0$ .

Equation of tangent to the circle at $(2 + \sqrt{3}, 3)$ is given by

$(2 + \sqrt{3}) x + 3 y - 2 (x + 2 + \sqrt{3}) - 4 (y + 3) + 16 = 0 \Rightarrow \sqrt{3} x - y - 2 \sqrt{3} = 0$

This line makes angle of $60^{\circ}$ with $x$ -axis. So initial position of the center is $A = (2, 4)$ and final position of the center is $B = (2 + 2 \cos 60^{\circ}, 4 + 2 \sin 60^{\circ}) = (3, 4 + \sqrt{3})$

Radius of the circle is $\sqrt{2^{2} + 4^{2} - 16} = 2$

Thus, equation of the circle in final position is $(x - 3)^{2} + {(y - 4 - \sqrt{3})}^{2} = 2^{2}$

$\Rightarrow x^{2} + y^{2} - 6 x - 2 (4 + \sqrt{3}) y + 24 + 8 \sqrt{3} = 0$ .

Let $(α, β)$ be the center and $a$ the radius of the required circle, then its equation is

$(x - α)^{2} + (y - β)^{2} = a^{2}$ . According to the questions $α^{2} + β^{2} = 1$

Since it touches $y = 0$ we have $| β | = a \Rightarrow b e t a = \pm a$

It also touches $y = \sqrt{3} (x + 1)$ we have $\frac{| \sqrt{3} α - β + \sqrt{3} |}{2} = a \Rightarrow \sqrt{3} α - β + \sqrt{3} = \pm 2 a$

Case I: $β = a$ and $\sqrt{3} α - β + \sqrt{3} = 2 a$ gives us $α = \sqrt{3} a - 1, β = a$

Thus, ${(\sqrt{3} a - 1)}^{2} + a^{2} = 1 \Rightarrow a = \frac{\sqrt{3}}{2}$

Hence, center of the circle is $(\frac{1}{2}, \frac{\sqrt{3}}{2})$ and radius is $\frac{\sqrt{3}}{2}$ .

Case II: $β = a$ and $\sqrt{3} α - β + \sqrt{3} = - 2 a$ gives us $α = - \frac{a}{\sqrt{3}} - 1$

Thus, ${(\frac{a}{\sqrt{3}} + 1)}^{2} + a^{2} = a \Rightarrow a = - \frac{\sqrt{3}}{2} [a > 0]$

Case III: $β = - a$ and $\sqrt{3} α - β + \sqrt{3} = 2 a$ gives us $α = \frac{a}{\sqrt{3}} - 1$

Thus, $(\frac{a}{\sqrt{3}} - 1) + a^{2} = 1 \Rightarrow a = \frac{\sqrt{3}}{2}$

Hence, center of circle is $(- \frac{1}{2}, - \frac{\sqrt{3}}{2})$ and radius is $\frac{\sqrt{3}}{2}$ .

Case IV: $β = - a$ and $\sqrt{3} α - β + \sqrt{3} = - 2 a$ gives us $α = \sqrt{3} a - 1$

Thus, ${(\sqrt{3} a + 1)}^{2} + a^{2} = 1 \Rightarrow a = 0, - \frac{\sqrt{3}}{2} [a > 0]$ .

Let $C (α, β)$ be the center of the circle and $a$ be its radius. According to question

$\frac{| 4 α - 3 β - 24 |}{5} = \frac{| 4 α + 3 β - 42 |}{5} = {\sqrt{(α - 2)^{2} + (β - 8)}}^{2}$

Solving these gives $β = 3, α = \frac{33}{4}$ but it is given that $| α | \leq 8$

Putting $β = 3$ in first and third one we get $α = 2, - \frac{182}{9} \Rightarrow α = 2$

Hence, equation of the circle is $(x - 2)^{2} + (y - 3)^{2} = (2 - 2)^{2} + (3 - 8)^{2} = 25$ .

Equation of the circle is $(x - 1)^{2} + (y + 5)^{2} = 7^{2} \Rightarrow x^{2} - 2 x + y^{2} + 10 y - 23 = 0$ .

Equation of the circle is $(x + 1)^{2} + (y + 2)^{2} = \frac{625}{4} \Rightarrow x^{2} + y^{2} + 2 x + 4 y + - \frac{605}{4} = 0$ .

Given diameters are $2 x + y = 6$ and $3 x + 2 y = 4$ . Solving we get $3 x + 2 (6 - 2 x) = 4 \Rightarrow x = 8, y = - 10$ .

Thus, equation of the circle is $(x - 8)^{2} + (y + 10)^{2} = 10^{2} \Rightarrow x^{2} + y^{2} - 16 x + 20 y + 64 = 0$ .

Given lines are $3 x - 2 y - 1 = 0$ and $4 x + y - 27 = 0$ . Thus, $y = 27 - 4 x$ and

$3 x - 2 (27 - 4 x) - 1 = 0 \Rightarrow x = 5, y = 7$ .

Let the radius of the circle is $r$ . Then equation will be $(x - 2)^{2} + (y - 3)^{2} = r^{2}$ .

Since it passes theough $(5, 7)$ , therefore, $3^{2} + 4^{2} = r^{2} \Rightarrow r = 5$

Thus, equation will be $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ .

Given lines are $3 x + y = 14$ and $2 x + 5 y = 18$ . From first $y = 14 - 3 x$ . Putting in the second $2 x + 5 (14 - 3 x) = 18 \Rightarrow x = 4 \Rightarrow y = 2$

Let the radius of the circle is $r$ , then the equation is $(x - 1)^{2} + (y - 2)^{2} = r^{2}$

Since it passes through $(4, 2)$ , we have $3^{2} + 0 = r^{2} \Rightarrow r = 3$ .

Thus, equation of the circle is $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ .

Given that the circle passes through $(2, 3)$ and has center $(1, 4)$ .

Let the raidus be $r$ , then equation of the circle is $(x - 1)^{2} + (y - 4)^{2} = r^{2}$ . Since it passes through $(2, 3)$

$(2 - 1)^{2} + (3 - 4)^{2} = r^{2} => r = \sqrt{2}$

Thus, equation of the circle is $(x - 1)^{2} + (y - 4)^{2} = 2 \Rightarrow x^{2} + y^{2} - 2 x - 8 y + 15 = 0$ .

Point of intersection of $x + 3 y = 0$ and $2 x - 7 y = 0$ is $(0, 0)$ . The other two lines are $x + y + 1 = 0$ and $x - 2 y + 4 = 0$ .

Thus, $y = - x - 1$ .Putting in $x - 2 y + 4 = 0 \Rightarrow x - 2 (- x - 1) + 4 = 0 \Rightarrow x = 2, y = - 3$ . So center is $(2, - 3)$ .

Let radius of the circle is $r$ . Then the equation of the circle is $(x - 2)^{2} + (y + 3)^{2} = r^{2}$

Since it passes through $(0, 0)$ , therefore, $r^{2} = 13$ .

Thus, equation of the circle is $x^{2} + y^{2} - 4 x + 6 y = 0$ .

Given that the radius is $5$ and center is at $(5, 0)$ .

Thus, equation of the circle is $(x - 5)^{2} + y^{2} = 5^{2} \Rightarrow x^{2} + y^{2} - 10 x = 0$ .

Let the center be $(h, k)$ . Since it lies on the line $x - 2 y = 0$ , we have $h = 2 k$ .

Because the circle passes through $(- 1, 2)$ and $(3, - 2)$ , the distances from the center to these points are equal. So $(h + 1)^{2} + (k - 2)^{2} = (h - 3)^{2} + (k + 2)^{2}$ .

Simplifying this gives $h - k = 1$ . Using $h = 2 k$ , we get $2 k - k = 1$ , so $k = 1$ and $h = 2$ .

Thus, the center is $(2, 1)$ . The radius squared is $(2 + 1)^{2} + (1 - 2)^{2} = 10$ .

Therefore, the equation of the circle is $(x - 2)^{2} + (y - 1)^{2} = 10$ .

Let the center be $(h, k)$ with $3 h + 4 k = 7$ .

Since the circle passes through $(1, - 2)$ and $(4, - 3)$ , we equate distances $(h - 1)^{2} + (k + 2)^{2} = (h - 4)^{2} + (k + 3)^{2}$ .

Expanding and simplifying gives $6 h - 2 k - 20 = 0$ , which reduces to $3 h - k = 10$ .

Now solve $3 h + 4 k = 7$ and $3 h - k = 10$ .

Subtracting gives $5 k = - 3$ , so $k = - \frac{3}{5}$ . Substitute into $3 h - k = 10 \Rightarrow 3 h + \frac{3}{5} = 10$ , so $3 h = \frac{47}{5}$ and $h = \frac{47}{15}$ .

Thus, the center is $(\frac{47}{15}, - \frac{3}{5})$ .

Now compute radius squared using $(1, - 2), {(\frac{47}{15} - 1)}^{2} + {(\frac{3}{5} + 2)}^{2} = {(\frac{32}{15})}^{2} + {(\frac{7}{5})}^{2} = \frac{1465}{225}$ .

So the equation of the circle is ${(x - \frac{47}{15})}^{2} + {(y + \frac{3}{5})}^{2} = \frac{1465}{225}$ .

Center of given circle is $(1, 2)$ and radius is $5$ .

Let required center be $(h, k)$ . Point $(5, 5)$ lies on it $\Rightarrow (h - 5)^{2} + (k - 5)^{2} = 25$

Collinearity gives $\frac{k - 5}{h - 5} = \frac{3}{4}$

So $k - 5 = \frac{3}{4} (h - 5)$

Substitute $(h - 5)^{2} + {(\frac{3}{4} (h - 5))}^{2} = 25 \Rightarrow h = 9$ or $1$

Reject $(1, 2)$ , so center is $(9, 8)$ .

Eqaution of the cicle is $(x - 9)^{2} + (y - 8)^{2} = 25$ .

Perpendicular distance from the center to the tangent will be radius. Let it be $r$ .

$r = \frac{| 2.1 - (- 3) - 4 |}{\sqrt{2^{2} + (- 1)^{2}}} = \frac{1}{\sqrt{5}}$

Thus, equation of the circle is $(x - 1) + (y + 3)^{2} = \frac{1}{5} \Rightarrow x^{2} + y^{2} - 2 x + 6 y + \frac{49}{5} = 0$ .

Since the circle is concentric with the circle $x^{2} + y^{2} - 4 x + 6 y - 17 = 0$ , therefore, center will be $(2, - 3)$ .

The circle touches the line $3 x - 4 y + 7 = 0$ , therefore, perpendicular distance from tangent will be radius. Let it be $r$ .

$r = \frac{| 3.2 - 4 (- 3) + 7 |}{\sqrt{3^{2} + (- 4)^{2}}} = 5$ .

Thus, the equation of the circle is $(x - 2)^{2} + (y + 3)^{2} = 5^{2} \Rightarrow x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ .

Let $(h, k)$ be the center then $(h - 1)^{2} + (k - 2)^{2} = 25$ .

The center is on normal perpendicular to give line. Hence slope will be $- \frac{4}{3}$ . Also, slope is $\frac{k - 2}{h - 1}$ .

Thus, $\frac{k - 2}{h - 1} = - \frac{4}{3} \Rightarrow 4 h + 3 k = 10 \Rightarrow k = \frac{10 - 4 h}{3}$ .

Putting this in the equation of the circle

$(h - 1)^{2} + {(\frac{10 - 4 h}{3} - 2)}^{2} = 25 \Rightarrow h^{2} - 2 h + 1 + \frac{16 h^{2} - 32 h + 16}{9} = 25 \Rightarrow 25 h^{2} - 50 h - 200 = 0$

$h^{2} - 2 h - 8 = 0 \Rightarrow h = 4, - 2, k = - 2, 6$ .

Thus, equation of the circles are $x^{2} + y^{2} - 8 x + 4 y - 5 = 0$ and $x^{2} + y^{2} + 4 x - 12 y + 15 = 0$ .

The two normals are $3 x - 5 y + 2 = 0$ and $x + 2 y = 3$ . Solving them $3 (3 - 2 y) - 5 y + 2 = 0 \Rightarrow y = 1 \Rightarrow x = 1$ .

This point of intersection of two normals give the center of the circle. Thus, the center is $(1, 1)$ , and it is given that radius of the circle is $5$ . Hence, the equation is

$(x - 1)^{2} + (y - 1)^{2} = 5^{2} \Rightarrow x^{2} + y^{2} - 2 x - 2 y - 23 = 0$ .

Let the center of the circle be $(a, 4)$ and radius be $r$ .

Since the circle touches the $y$ -axis at the point $(0, 4)$ , the radius equals the distance of the center from the $y$ -axis.

Hence $r = a$ .

Equation of the circle $(x - a)^{2} + (y - 4)^{2} = a^{2}$

The circle cuts an intercept $6$ on the $x$ -axis.

So the points where the circle meets the $x$ -axis satisfy $y = 0$ .

Substituting $y = 0, (x - a)^{2} + 16 = a^{2} \Rightarrow (x - a)^{2} = a^{2} - 16$

The length of the intercept on the $x$ -axis is $2 \sqrt{a^{2} - 16} = 6 \Rightarrow a = \pm 5$

Hence, the radius is $5$ and the center is $(\pm 5, 4)$ .

Therefore, the equation of the circle is $(x \mp 5)^{2} + (y - 4)^{2} = 25$ .

The circle would be in any quadrant and center would be one of
$(a, a), (- a, a), (- a, - a)$ and $(a, - a)$ .

Thus, equation of the circle is $(x \mp a)^{2} + (y \mp a)^{2} = a^{2}$ .

Since the circle touches the $y$ -axis at origin, therefore, the $x$ -axis will be normal. If $r$ is the radius then $(r, 0)$ will be the center.

Since it passes through $(h, k)$ , therefore, $(h - r)^{2} + k^{2} = r^{2} \Rightarrow h^{2} + k^{2} - 2 h r = 0 \Rightarrow r = \frac{h^{2} + k^{2}}{2 h}$

Thus, equation of the circle is $(x - r)^{2} + y^{2} = r^{2} \Rightarrow x^{2} + y^{2} - 2 r x = 0 \Rightarrow h (x^{2} + y^{2}) - (h^{2} + k^{2}) x = 0$ .

Since the circle touches the $x$ -axis at origin, therefore, the $y$ -axis will be normal. If $r$ is the radius then $(0, r)$ will be the center.

The circle touches the line $4 x - 3 y + 24 = 0$ , so perpendicular distance from the center will be equal to the radius.

$\frac{| 4.0 - 3 r + 24 |}{5} = r \Rightarrow 9 (r - 8)^{2} = 25 r^{2} \Rightarrow 9 r^{2} - 144 r + 576 = 25 r^{2} \Rightarrow 16 r^{2} + 144 r - 576 = 0$

$\Rightarrow r^{2} + 9 r - 36 = 0 \Rightarrow r = 3, - 12$

Thus, equation of the circles are $x^{2} + y^{2} + 24 y = 0$ and $x^{2} + y^{2} - 6 y = 0$ .

Given circle is $x^{2} + y^{2} = 16 = 4^{2}$ . Thus, parametric equaiton is $x = 4 \cos θ, y = 4 \sin θ$ .

Let the center of the circle be $(h, k)$ and radius be $r$ .

The circle touches the line $2 x - y = 1$ at the point $(1, 1)$ .

Hence, the center lies on the line perpendicular to $2 x - y = 1$ through $(1, 1)$ .

Slope of $2 x - y = 1$ is $2$ . Therefore the perpendicular slope is $- \frac{1}{2}$ .

Equation of the perpendicular line through $(1, 1), y - 1 = (- \frac{1}{2}) (x - 1) \Rightarrow x + 2 y - 3 = 0$

So the center $(h, k)$ satisfies $h + 2 k - 3 = 0$

The radius equals the distance from the center to the line $2 x - y - 1 = 0$ and also to the line $2 x + y - 4 = 0$

Hence, $\frac{| 2 h - k - 1 |}{\sqrt{5}} = \frac{| 2 h + k - 4 |}{\sqrt{5}} \Rightarrow | 2 h - k - 1 | = | 2 h + k - 4 |$

Using $h + 2 k - 3 = 0$ , we get $h = 3 - 2 k$

Substituting in the distance equation $| 2 (3 - 2 k) - k - 1 | = | 2 (3 - 2 k) + k - 4 | \Rightarrow k = \frac{7}{8}$

Then $h = 3 - 2 \frac{7}{8} \Rightarrow h = \frac{5}{4}$

Radius $r = \frac{| 2 \frac{5}{4} - \frac{7}{8} - 1 |}{\sqrt{5}} \Rightarrow r = \frac{\sqrt{5}}{8}$

Therefore, the equation of the circle is ${(x - \frac{5}{4})}^{2} + {(y - \frac{7}{8})}^{2} = \frac{5}{64}$ .

Let the center be $(h, k)$ and radius be $r$ . Since the center lies on $2 x + y = 0$ , therefore, $k = - 2 h$

The circle touches the lines $4 x - 3 y - 30 = 0$ and $4 x - 3 y + 10 = 0$

Hence, the distances from the center to both lines are equal.

$\frac{| 4 h - 3 k - 30 |}{5} = \frac{| 4 h - 3 k + 10 |}{5} \Rightarrow 4 h - 3 k = 10$

Using $k = - 2 \Rightarrow 4 h + 6 h = 10 \Rightarrow h = 1, k = - 2$

Radius $r = \frac{| 4 (1) - 3 (- 2) - 30 |}{5} = 4$

Therefore, the equation of the circle is $(x - 1)^{2} + (y + 2)^{2} = 16$ .

Since the circle touches the coordinate axes and has a radius of $4$ units, therefore, its center would be $(4, 4)$ in first quadrant.

Given that $y = 0$ is the line mirror, so the new center would be $(4, - 4)$ and the equation of circle will be

$(x - 4)^{2} + (y + 4)^{2} = 16$ .

Figure 11.4.

Center of the circle $x^{2} + y^{2} + 16 x - 24 y + 183 = 0$ is $(- 8, 12)$ and radius is $\sqrt{64 + 144 - 183} = 5$ .

Figure 11.5.

Image of the center across this line is given by $x^{'} = x - \frac{2 a (a b + b y + c)}{a^{2} + b^{2}}$ and $y^{'} = y - \frac{2 b (a x + b y + c)}{a^{2} + b 62}$

Here $a = 4, b = 7, c = 13$ and $(x, y) = (- 8, 12)$ .

Now $a x + b y + c = 4 (- 8) + 7 (12) + 13 = - 32 + 84 + 13 = 65$

Also $a^{2} + b^{2} = 16 + 49 = 65$ . Therefore $x^{'} = - 8 - (2.4 .65) / 65 = - 16$ and $y^{'} = 12 - (2.7 .65) / 65$ $= - 2$

Hence the image circle has center $(- 16, - 2)$ and radius $5$ .

Thereforem its equation is $(x + 16)^{2} + (y + 2)^{2} = 25 \Rightarrow x^{2} + y^{2} + 32 x + 4 y + 235 = 0$ .

Center of the circle is $(a, a)$ and radius is $a$ . After one complete revolution along $x$ -axis will make the new center as $(a + 2 π a, a)$ .

Thus, new equation is $(x - a - 2 π a)^{2} + (y - a)^{2} = a^{2}$ .

The center is $(1, 1)$ and radius is $5$ . The center moves on the line $x - y = 0$ , so any new center $(h, k)$ satisfies

$h - k = 0 \Rightarrow h = k$

Distance between old and new center is $\sqrt{2}$ , so $(h - 1)^{2} + (k - 1)^{2} = 2$

Substitute $k = h$ $(h - 1)^{2} + (h - 1)^{2} = 2$

$2 (h - 1)^{2} = 2 \Rightarrow h - 1 = \pm 1$

Thus two circles are possible. $h = 2, k = 2$ and $h = 0, k = 0$

For $(2, 2)$ equation of circle will be $(x - 2)^{2} + (y - 2)^{2} = 25$ .

For $(0, 0)$ equation of circle will be $x^{2} + y^{2} = 25$ .

Let the circle pass through the origin and have center $(h, k)$ and radius $r$ .

Since it passes through $(0, 0), h^{2} + k^{2} = r^{2}$

The length of chord cut by the circle on a line is $2 \sqrt{r^{2} - d^{2}}$ where $d$ is perpendicular distance from center to the line.

Given chord length is $\sqrt{2}$ .

So $2 \sqrt{r^{2} - d^{2}} = \sqrt{2} \Rightarrow r^{2} - d^{2} = \frac{1}{2} \Rightarrow h^{2} + k^{2} - d^{2} = 1 / 2$

Now for line $y = x \Rightarrow x - y = 0$

Distance from $(h, k)$ is $d_{1} = \frac{| h - k |}{\sqrt{2}}$

So $h^{2} + k^{2} - \frac{(h - k)^{2}}{2} = \frac{1}{2}$

Simplifying gives us $(h + k)^{2} = 1 \dots$ (1)

Now for line $y = - x \Rightarrow x + y = 0$ . Distance is $d_{2} = \frac{| h + k |}{\sqrt{2}}$

So similarly $h^{2} + k^{2} - \frac{(h + k)^{2}}{2} = \frac{1}{2}$

Simplifying gives $(h - k)^{2} = 1 \dots$ (2)

From (1) and (2): $h + k = \pm 1$ and $h - k = \pm 1$

Case I: $h + k = 1, h - k = 1 \Rightarrow h = 1, k = 0$

Case II: $h + k = 1, h - k = - 1 \Rightarrow h = 0, k = 1$

Case III: $h + k = - 1, h - k = 1 \Rightarrow h = 0, k = - 1$

Case IV: $h + k = - 1, h - k = - 1 \Rightarrow h = - 1, k = 0$

Radius: $r^{2} = h^{2} + k^{2} = 1$ . So $r = 1$

Thus the circles are $(x - 1)^{2} + y^{2} = 1, x^{2} + (y - 1)^{2} = 1, x^{2} + (y + 1)^{2} = 1$ , and $(x + 1)^{2} + y^{2} = 1$ .

Let $A = 3 x + 4 y - 15 = 0, B = 3 x - 4 y - 7 = 0, C = 12 x + 5 y - 115 = 0$

$\frac{| 3 x + 4 y - 15 |}{5} = \frac{3 x - 4 y - 7 |}{5} \Rightarrow 3 x + 4 y - 15 = \pm (3 x - 4 y - 7)$

Taking plus sign $3 x + 4 y - 15 = 3 x - 4 y - 7 \Rightarrow 8 y = 8 \Rightarrow y = 1$

Now use the bisector of $A$ and $C$ , $\frac{| 3 x + 4 y - 15 |}{5} = \frac{| 12 x + 5 y - 115 |}{13}$

Substitute $y = 1$ ot get $\frac{| 3 x - 11 |}{5} = \frac{| 12 x - 110 |}{13} \Rightarrow x = 7$

Hence the incenter is $(7, 1)$ .

Let the raidus be $r$ . The center could be in any quadrant, and thus, center is $(\pm r, \pm r)$ .

Since the center lies on $l x + m y + n = 0$ , therefore, $r = \pm \frac{n}{l + m}$ .

Putting this is in $(x \pm r)^{2} + (y \pm r)^{2} = r^{2}$ gives us

$(l \pm m)^{2} (x^{2} + y^{2}) \pm 2 n (l \pm m) (x + y) + n^{2} = 0$ .

The smaller circle is $x^{2} + y^{2} = 4$ so its centre is $(0, 0)$ and radius is $2$ .

Let the radius of the larger circle be $R$ . The distance of the line $x + y = 2$ from the centre is $\sqrt{2}$

Chord length in a circle of radius $r$ at distance $d$ from the centre is $2 \sqrt{r^{2} - d^{2}}$

Hence, the chord lengths are $2 \sqrt{4 - 2} = 2 \sqrt{2}$ and $2 \sqrt{R^{2} - 2}$

Given intercept between the circles is $1$

$\Rightarrow \sqrt{R^{2} - 2} - \sqrt{2} = 1 \Rightarrow \sqrt{R^{2} - 2} = 1 + \sqrt{2} \Rightarrow R^{2} = 5 + 2 \sqrt{2}$

Therefore, the larger circle is $x^{2} + y^{2} = 5 + 2 \sqrt{2}$ .

The given circle is $x^{2} + y^{2} - 2 x + y = 0$ . So the centre is $(1, - \frac{1}{2})$ and radius is $\frac{\sqrt{5}}{2}$ .

For the point $(3, 2)$ , the distance from the centre is $\sqrt{(3 - 1)^{2} + (2 + 1 / 2)^{2}} = \frac{\sqrt{41}}{2}$

Since $\sqrt{41} 2 > \frac{\sqrt{5}}{2}$ the point is exterior to the circle.

The maximum radius of a circle centered at $(3, 2)$ containing the given circle is

$\frac{\sqrt{41}}{2} + \frac{\sqrt{5}}{2} = \frac{\sqrt{41} + \sqrt{5}}{2}$

Hence the required circle is $(x - 3)^{2} + (y - 2)^{2} = {(\frac{\sqrt{41} + \sqrt{5}}{2})}^{2} \Rightarrow (x - 3)^{2} + (y - 2)^{2} = \frac{23 + \sqrt{205}}{2}$ .

$C_{1} = (0, 2), r_{1} = 3, C_{2} = (- 6, - 2), r_{2} = 3$ , and $C_{3} = (- 3, - 6), r_{3} = 3$

$Δ C_{1} C_{2} C_{3}$ is acute, so the minimum enclosing circle of the centers is the circumcircle

$C = (- \frac{31}{18}, - \frac{23}{12}) \Rightarrow R = \frac{5 \sqrt{949}}{36}$ , where $C$ is circumcenter.

Since each given circle has radius $3$ , the required minimum radius is $R_{m i n} = 3 + \frac{5 \sqrt{949}}{36} = \frac{108 + 5 \sqrt{949}}{36}$

Hence, the required circle is ${(x + \frac{31}{18})}^{2} + {(y + \frac{23}{12})}^{2} = {(\frac{108 + 5 \sqrt{949}}{36})}^{2}$ .

The diamter form of the circle is $(x + 4) (x - 12) + (y - 3) (y + 1) = 0 \Rightarrow x^{2} + y^{2} - 8 x - 2 y - 51 = 0$ .

Putting $x = 0$ for intercept on $y$ -axis, we have $y^{2} - 2 y - 51 = 0$

Let $y_{1}$ and $y_{2}$ be the roots then $y_{1} + y_{2} = 2$ and $y_{1} y_{2} = - 51$ , then

$| y_{1} - y_{2} | = \sqrt{(y_{1} + y_{2})^{2} - 4 y_{1} y_{2}} = 4 \sqrt{13}$ .

So the intercept on $y$ -axis is $4 \sqrt{13}$ .

One of the diagonals will have endpoints as $(1, 2)$ and $(3, 4)$ and the other will have $(3, 2)$ and $(1, 4)$ .

Thus, the equation of the circle is $(x - 1) (x - 3) + (y - 2) (y - 4) = 0$ .

The other diagonal will also give the same equation.

Equation of the circle will be $x (x - 2) + y (y + 4) = 0$ .

Equation of the circle will be $(x - 2) (x + 2) + (y + 3) (y - 4) = 0 \Rightarrow x^{2} - 4 + y^{2} - y - 12 = 0$

So center will be $(0, \frac{1}{2})$ and radius will be $\sqrt{\frac{1}{4} + 16} = \frac{\sqrt{65}}{2}$ .

The intercepts are $(3, 0)$ and $(0, 4)$ on $x$ and $y$ axes. Thus, the equation of the circle will be

$x (x - 3) + y (y - 4) = 0$ .

The center of the circles are $(- 3, 7)$ and $(2, - 5)$ . Thus, the equation of the circle is

$(x + 3) (x - 2) + (y - 7) (y + 5) = 0$ .

One of the diagonals will have endpoints as $(6, 3)$ and $(9, 6)$ . Thus equation of the circle will be

$(x - 6) (x - 9) + (y - 3) (y - 6) = 0$ . The other diagonal will also give the same equation.

The given lines form a rectangle since there are two pairs of parallel lines.

From $x - 3 y = 4$ and $3 x + y = 22$ we get point $A (7, 1)$ .

From $x - 3 y = 14$ and $3 x + y = 22$ we get point $B (8, - 2)$ .

From $x - 3 y = 14$ and $3 x + y = 62$ we get point $C (20, 2)$ .

From $x - 3 y = 4$ and $3 x + y = 62$ we get point $D (19, 5)$ .

We take one diagonal $(7, 1)$ and $(20, 2)$ as diameter to get the equation of the circle as

$(x - 7) (x - 20) + (y - 1) (y - 2) = 0$ . The other diagonal will also give the same equation.

Let $x_{1}, x_{2}$ be the roots of $x^{2} + 2 x - a^{2} = 0$ then $x_{1} + x_{2} = - 2$ and $x_{1} x_{2} = - a^{2}$ . Similarly, let $y_{1}, y_{2}$ be the roots of $y^{2} + 4 y - b^{2} = 0$ then $y_{1} + y_{2} = - 4$ and $y_{1} y_{2} = - b^{2}$ .

Circle whose endpoints will be the diamter $A B$ will be given by $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

$\Rightarrow x^{2} - x (x_{1} + x_{2}) + x_{1} x_{2} + y^{2} - y (y_{1} + y_{2}) + y_{1} y_{2} = 0$

Substituting the values from the equations obtained we have the equation as

$x^{2} + 2 x - a^{2} + y^{2} + 4 y - b^{2} = 0 = (x + 1)^{2} + (y + 2)^{2} = a^{2} + b^{2} + 5$

Hence, center is $(- 1, - 2)$ and radius is $\sqrt{a^{2} + b^{2} + 5}$ .

The circle is given by $x^{2} + y^{2} - 2 x + 6 y - 15 = 0$ . Let $(h, k)$ be the other endpoint. Then the equation for the circle with the diameter is given by

$(x - 4) (x - h) + (y - 1) (y - k) = 0 \Rightarrow x^{2} - (4 + h) + 4 h + y^{2} - (1 + k) + k = 0$

Comparing coefficients of $x$ and $y$ we have $h = - 2$ and $k = - 7$ .

The given lines are $a x + b y + c = 0, a x + b y - c = 0, b x - a y + c = 0$ and $b x - a y - c = 0$ .

These form a rectangle since each pair is parallel and the two directions are perpendicular.

Take one pair of opposite vertices by solving $a x + b y + c = 0$ with $b x - a y + c = 0$ and $a x + b y - c = 0$ with $b x - a y - c = 0$ .

Let these points be $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ .

From symmetry we have $x_{2} = - x_{1}$ and $y_{2} = - y_{1}$ . So the equation of the circle becomes $x^{2} + y^{2} = x_{1}^{2} + y_{1}^{2}$ .

Solving $a x + b y + c = 0$ and $b x - a y + c = 0$ gives $x_{1} = - \frac{c (a + b)}{a^{2} + b^{2}}$ and $y_{1} = - \frac{c (b - a)}{a^{2} + b^{2}}$ .

So $x_{1}^{2} + y_{1}^{2} = \frac{c^{2} [(a + b)^{2} + (b - a)^{2}]}{{(a^{2} + b^{2})}^{2}}$

This simplifies to $x_{1}^{2} + y_{1}^{2} = \frac{2 c^{2}}{a^{2} + b^{2}}$ .

Hence, the equation of the circumcircle is $x^{2} + y^{2} = \frac{2 c^{2}}{a^{2} + b^{2}}$ .

Let the equation of the circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ whose center is $(- g, - f)$ , which lies on $3 x + 4 y = 7$ . Thus,

$- 3 g - 4 f = 7$ . Since the circle passes through $(1, - 2)$ and $(4, - 3)$ , therefore,

$2 g - 4 f + c = - 5$ and $8 g - 6 f + c = 25$ . From these two equations we have $- 3 g + f = 10$

Thus, $f = \frac{3}{5}, g = - \frac{47}{15}$ and $c = \frac{11}{3}$ and now it is trivial to find the equation.

The line $3 x + 4 y = 12$ meets the axes at $(4, 0)$ and $(0, 3)$ . Let the equation of the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Since it passes through origin, therefore, $c = 0$ .

For $(4, 0)$ the equation becomes $16 + 8 g = 0 \Rightarrow g = - 2$ and for $(0, 3)$ the equation is $9 + 6 f = 0 \Rightarrow f = - \frac{3}{2}$ . Thus, we have found the equation of the circle as $x^{2} + y^{2} - 4 x - 3 y = 0$ .

The given lines are $5 x + 3 y = 9, x = 3 y, 2 x = y$ and $x + 4 y + 2 = 0$ .

Figure 11.6.

From $5 x + 3 y = 9$ and $x = 3 y$ we get $A (\frac{3}{2}, \frac{1}{2})$ .

From $x = 3 y$ and $2 x = y$ we get $B (0, 0)$ .

From $2 x = y$ and $x + 4 y + 2 = 0$ we get $C (- \frac{2}{9}, - \frac{4}{9})$ .

We take the general circle $x^{2} + y^{2} + g x + f y + c = 0$ .

Substitute point $B (0, 0)$ and get $c = 0$ .

Substitute point $A (\frac{3}{2}, \frac{1}{2})$ .

This gives $\frac{5}{2} + \frac{3 g}{2} + \frac{f}{2} = 0$ so $3 g + f = - 5$ .

Substitute point $C (- 2 / 9, - 4 / 9)$ .

This gives $\frac{20}{81} - \frac{2 g}{9} - \frac{4 f}{9} = 0$ .

Multiply by $81$ to get $20 - 18 g - 36 f = 0$ so $9 g + 18 f = 10$ .

Thus, $g = - \frac{20}{9}$ and $f = \frac{5}{3}$ .

Hence, the circle is $x^{2} + y^{2} - \frac{20}{9} x + \frac{5}{3} y = 0$ .

Let the equation of the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Since it passes through $(1, 2)$ and $(3, 4)$ , therefore,

$5 + 2 g + 4 f + c = 0 \dots$ (1) and $25 + 6 g + 8 f + c = 0 \dots$ (2)

From these two equations we have $g + f + 5 = 0) \dots$ (3)

Since the circle touches the line $3 x + y - 3 = 0$ so perpendicular distance from center would be radius. Thus,

$\frac{| - 3 g - f - 3 |}{\sqrt{10}} = \sqrt{g^{2} + f^{2} + c} \Rightarrow (3 g + f + 3)^{2} = 10 (g^{2} + f^{2} + 5 + 2 g + 4 f)$ [from (1)]

$(2 g - 5 + 3)^{2} = 10 [g^{2} + (g + 5)^{2} + 5 + 2 g - 4 g - 29]$ [Putting the value of $f$ from (3)]

$\Rightarrow g = - 4, - \frac{3}{2} \Rightarrow f = - 1, - \frac{7}{2} \Rightarrow c = 7, 12$

Thus, equation of the circles are $x^{2} + y^{2} - 8 x - 2 y + 7 = 0$ and $x^{2} + y^{2} - 3 x - 7 y + 12 = 0$ .

Let the equation of the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , so the center is $(- g, - f)$ and radius $\sqrt{g^{2} + f^{2} - c}$

Since the circle touches the $x$ -axis, therefore, $g^{2} - c = 0 \Rightarrow c = g^{2} [∵ \sqrt{g^{2} + f^{2} - c} = | - f |] \dots$ (1)

Also, the circle touches $4 x - 3 y + 4 = 0 \Rightarrow \frac{| - 4 g + 3 f + 4 |}{5} = \sqrt{g^{2} + f^{2} + c} = | f |$ [from (1)]

$\Rightarrow - 4 g + 3 f + 4 = \pm f \Rightarrow 2 g + f = 2 \dots$ (2) and $g - 2 f = 1 \dots$ (3)

Also, given that the center lies on $x - y - 1 = 0 \Rightarrow - g + f = 1 \dots$ (4)

Thus, $g = \frac{1}{3}, f = \frac{4}{3}$ and $f = - 2, g = - 3$ , which lies in first quadrant. Thus, $c = \frac{1}{9}$ .

Hence, the equation of the cirlce is $9 (x^{2} + y^{2}) + 6 x + 24 y + 1 = 0$ .

Let the circle be $x^{2} + y^{2} + g x + f y + c = 0$ . Substituting point $(1, 0)$ gives

$1 + g + c = 0$ so $g + c = - 1$ .

Substituting point $(0, 1)$ gives $1 + f + c = 0$ so $f + c = - 1$ .

Substituting point $(1, - 2)$ gives $1 + 4 + g - 2 f + c = 0$ so $g - 2 f + c = - 5$ .

From $g + c = - 1$ we get $g = - 1 - c$ . From $f + c = - 1$ we get $f = - 1 - c$ .

Substitute into $g - 2 f + c = - 5$ .

This gives $(- 1 - c) - 2 (- 1 - c) + c = - 5$ . So $- 1 - c + 2 + 2 c + c = - 5$ .

This simplifies to $1 + 2 c = - 5$ so $c = - 3$ . Then $g = 2$ and $f = 2$ .

Hence, the equation of the circle is $x^{2} + y^{2} + 2 x + 2 y - 3 = 0$ .

Let the circle be $x^{2} + y^{2} + g x + f y + c = 0$ .

Substituting point $(0, 0)$ gives $c = 0$ .

Substituting point $(a, 0)$ gives $a^{2} + g a = 0$ so $g = - a$ .

Substituting point $(0, b)$ gives $b^{2} + f b = 0$ so $f = - b$ .

Hence, the equation of the circle is $x^{2} + y^{2} - a x - b y = 0$ .

Let the circle be $x^{2} + y^{2} + g x + f y + c = 0$ . Since it passes through the origin, we get $c = 0$ .

So the equation becomes $x^{2} + y^{2} + g x + f y = 0$ .

Now we consider the intercept on the positive $x$ -axis. Putting $y = 0$ gives $x^{2} + g x = 0$ which gives $x (x + g) = 0$ .

So the intercept points are $x = 0$ and $x = - g$ .

The length of the chord on the positive $x$ -axis is $4$ . Hence, $- g = 4$ so $g = - 4$ .

Now we consider the intercept on the positive $y$ -axis. Putting $x = 0$ gives $y^{2} + f y = 0$ which gives $y (y + f) = 0$ .

So the intercept points are $y = 0$ and $y = - f$ . The length of the chord on the positive $y$ -axis is $6$ .

Hence, $- f = 6$ so $f = - 6$ . Therefore, the equation of the circle is $x^{2} + y^{2} - 4 x - 6 y = 0$ .

The given lines are $y = x, y = 2 x$ and $y = 3 x + 2$ .

From $y = x$ and $y = 2 x$ we get $A (0, 0)$ .

From $y = x$ and $y = 3 x + 2$ we get $B (- 1, - 1)$ .

From $y = 2 x$ and $y = 3 x + 2$ we get $C (- 2, - 4)$ .

Let the circle be circle $x^{2} + y^{2} + g x + f y + c = 0$ .

Substituting point $A (0, 0)$ and get $c = 0$ .

Substituting point $B (- 1, - 1)$ gives $1 + 1 - g - f = 0$ so $g + f = 2$ .

Substituting point $C (- 2, - 4)$ gives $4 + 16 - 2 g - 4 f = 0$ so $g + 2 f = 10$ .

From $g + f = 2$ we get $g = 2 - f$ .

Substituting into $g + 2 f = 10$ gives $2 - f + 2 f = 10$ so $f = 8$ .

Then $g = - 6$ . Hence, the equation of the circumcircle is $x^{2} + y^{2} - 6 x + 8 y = 0$ .

The given sides of the triangle are $7 x - y + 11 = 0, x + y - 15 = 0$ and $7 x + 17 y + 65 = 0$ .

From $7 x - y + 11 = 0$ and $x + y - 15 = 0$ we get $A (\frac{1}{2}, \frac{29}{2})$ .

From $x + y - 15 = 0$ and $7 x + 17 y + 65 = 0$ we get $B (- 10, 25)$ .

From $7 x + 17 y + 65 = 0$ and $7 x - y + 11 = 0$ we get $C (- \frac{19}{3}, - \frac{100}{3})$ .

If the side lengths opposite $A, B, C$ are $a, b, c$ then the incenter is

$\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}$ and $\frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}$ .

Length $a = B C = 5 \sqrt{85}$ . Length $b = C A = 43 \sqrt{5}$ . Length $c = A B = 3 \sqrt{85}$ .

After simplification the incenter is $(- 3, 11)$ .

Now find the radius which is the perpendicular distance from the incenter to any side.

Distance to $x + y - 15 = 0$ is $\frac{| - 3 + 11 - 15 |}{\sqrt{2}} = \frac{7}{\sqrt{2}}$ .

Hence, the equation of the incircle is $(x + 3)^{2} + (y - 11)^{2} = \frac{49}{2}$ .

Let the circle be $x^{2} + y^{2} + g x + f y + c = 0$ . Since it passes through the origin, we get $c = 0$ .

So the equation becomes $x^{2} + y^{2} + g x + f y = 0$ .

Now we consider the line $3 x = 4 y$ which is $3 x - 4 y = 0$ .

The perpendicular distance from the center $(- \frac{g}{2}, - \frac{f}{2})$ to this line is

$\frac{| 3 (- \frac{g}{2}) - 4 (- \frac{f}{2}) |}{5} = \frac{| - 3 g + 4 f |}{10}$ .

Since the circle cuts off a chord of length $1$ on this line, we use

$1 = 2 \sqrt{r^{2} - d^{2}}$ . So $r^{2} - d^{2} = \frac{1}{4}$ .

Now $r^{2} = \frac{g^{2} + f^{2}}{4}$ . So $\frac{g^{2} + f^{2}}{4} - \frac{(- 3 g + 4 f)^{2}}{100} = \frac{1}{4}$ .

Similarly for the line $4 x = 3 y$ which is $4 x - 3 y = 0$ . Distance from center is $\frac{| - 4 g + 3 f |}{10}$ .

So $\frac{g^{2} + f^{2}}{4} - \frac{(- 4 g + 3 f)^{2}}{100} = \frac{1}{4}$ .

Now subtract the two equations. This gives $(- 3 g + 4 f)^{2} = (- 4 g + 3 f)^{2}$ .

So either $- 3 g + 4 f = - 4 g + 3 f$ or $- 3 g + 4 f = 4 g - 3 f$ . First case gives $g + f = 0$ . Second case gives $g = f$ .

Now substitute each case. For $g + f = 0$ we get $g = - f$ .

Substitute into equation and solve to get $g = 1$ and $f = - 1$ .

For $g = f$ we substitute and get $g = - 1$ and $f = - 1$ .

Hence the required circles are $x^{2} + y^{2} + x - y = 0$ and $x^{2} + y^{2} - x - y = 0$ .

Common chord of the circles is $x^{2} + y^{2} - 4 x - 5 - (x^{2} + y^{2} + 8 y + 7) = 0 \Rightarrow x + 2 y + 3 = 0$

Equation of such a circle is $x^{2} + y^{2} - 4 x - 5 + k (x + 2 y + 3) = 0 \Rightarrow x^{2} + y^{2} - (4 - k) x + 2 k y - 3 k - 5 = 0$ .

Its center is $(\frac{4 - k}{2}, - k)$ . If $x + 2 y + 3 = 0$ is diameter then $\frac{4 - k}{2} - 2 k + 3 = 0 \Rightarrow k = 2$ .

Thus, equation of the circle is $x^{2} + y^{2} - 2 x + 4 y + 1 = 0$ .

Equation of any circle passing through the point of intersection of the given circle and the given chord is $x^{2} + y^{2} - a^{2} + k (x \cos α + y \sin α - p) = 0$ .

Center of this circle is $(- \frac{k \cos α}{2}, - \frac{k \sin α}{2})$ .

Since $x \cos α + y \sin α - p = 0$ is the diameter of this circle the center will lie on this line, therefore,

$- \frac{k \cos α}{2} \cos α - \frac{k \sin α}{2} \sin α - p = 0 \Rightarrow k = - 2 p$

Thus, the equation of the circle becomes $x^{2} + y^{2} - a^{2} - 2 p (x \cos α + y \sin α - p) = 0$ .

Clearly $x^{2} + y^{2} - 4 = 0$ is the equation of a circle with center at origin and radius $2$ .

Also line $y = m x + 2 \sqrt{1 + m^{2}}$ is the equation of the line which touches the circle for all values for $m$ .

Let $P$ be the point of contact of the circle and the line. Clearly, given equation is the equation of circles passing through the point of contact of the given circle and the given line. Any two circles of this family touch each other at $P$ .

Equation of the line joining the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$

Also equation of the circle with $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ as endpoints of the diameter is $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$ .

Equation of any circle passing through the point of intersection of the above circle and line is given by $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) + λ | \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} | = 0$ .

Putting $λ = 0$ gives $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$ , which is the diameter form of the equation of the circle.

Equation of any circle through the point of intersection $P$ and $Q$ of the line and the circle is $x^{2} + y^{2} + a x + b y + c + λ (A x + B y + C) = 0$

$\Rightarrow x^{2} + y^{2} + (a + λ A) x + (b + λ B) y + c + λ C = 0$ .

Similarly for other pair of line and circle $x^{2} + y^{2} + (a^{'} + μ A^{'}) x + (b^{'} + μ B^{'}) y + c^{'} + μ C^{'} = 0$

If the two circles are the same then the points $P, Q, R$ and $S$ will be concyclic.

Comparing coefficients $1 = \frac{a + λ A}{a^{'} + μ A^{'}} = f r a c b + λ B b^{'} + μ B^{'} = \frac{c + λ C}{c^{'} + μ C^{'}}$

Thus, $a - a^{'} + λ A - m u A^{'} = 0, b - b^{'} + λ B - μ B^{'} = 0$ , and $c - c^{'} + λ C - μ C^{'} = 0$ .

Eliminating $λ$ and $- μ$ and writing in discriminant form we have

$| \begin{matrix} a - a^{'} & A & A^{'} \\ b - b^{'} & B & B^{'} \\ c - c^{'} & C & C^{'} \end{matrix} | = | \begin{matrix} a - a^{'} & b - b^{'} & c - c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{matrix} | = 0$

Equation of any circle possing through the points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ is given by

$S = (x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) + λ | \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} | = 0 \dots$ (1)

Let the fixed circle be $S^{'} = x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \dots$ (2)

Equation of the chord of intersection of circles (1) and (2) will be $S - S^{'} = 0$

$\Rightarrow - (x_{1} + x_{2} + 2 g) x - (y_{1} + y_{2} + 2 f) y + x_{1} x_{2} + y_{1} y_{2} - c + λ | \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} | = 0 \dots$ (3)

Clearly this line passes through the point of intersection of two fixed lines $- (x_{1} + x_{2} + 2 g) x - (y_{1} + y_{2} + 2 f) y + x_{1} x_{2} + y_{1} y_{2} - c = 0$ and $| \begin{matrix} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{matrix} | = 0$ , which is a fixed point.

Given circle is $x^{2} + y^{2} - a^{2} = 0 \dots$ (1)

Since $P Q$ and $P R$ are tangents to the circle (1), therefore $Q R$ will be the chord of contact of point $(x_{1}, y_{1})$ , and hence, equaiton of $Q R$ will be

$x x_{1} + y y_{1} - a^{2} = 0 \dots$ (2)

Equation of any circle through the point of intersection $Q$ and $R$ of (1) and (2) is

$x^{2} + y^{2} - a^{2} + k (x x_{1} + y y_{1} - a^{2}) = 0 \dots$ (3)

Circle (3) will be circumcircle of $△ P Q R$ if circle (3) passes through the point $P (x_{1}, y_{1})$ i.e.

$x_{1}^{2} + y_{1}^{2} - a^{2} + k (x_{1}^{2} + y_{1}^{2} - a^{2}) = 0 \Rightarrow k = - 1$

Hence, required circle is $x^{2} + y^{2} - x x_{1} - y y_{1} = 0$ .

Given circles are $x^{2} + y^{2} - 6 x + 2 y + 4 = 0 \dots$ (1) and $x^{2} + y^{2} + 2 x - 4 y - 6 = 0 \dots$ (2) and given line is $x - y = 0 \dots$ (3).

Equation of any circle passing through the point of intersection of circles (1) and (2) is

$x^{2} + y^{2} - 6 x + 2 y + 4 + k (x^{2} + y^{2} + 2 x - 4 y - 6) = 0 \Rightarrow (1 + k) x^{2} + (1 + k) y^{2} - 2 (3 - k) x + 2 (1 - 2 k) y + 4 - 6 k = 0$

Its center is $(\frac{3 - k}{1 + k}, \frac{2 k - 1}{1 + k})$ . Since it lies on the line (3), therefore,

$\frac{3 - k}{1 + k} - \frac{2 k - 1}{1 + k} = 0 \Rightarrow k = \frac{4}{3}$ .

Thus, required equation is $x^{2} + y^{2} - \frac{10}{7} x - \frac{10}{7} y - \frac{12}{7} = 0$ .

Let $S_{1} = x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \dots$ (1) and $S_{2} = x^{2} + y^{2} + 2 g^{'} x + 2 f^{'} y + c^{'} = 0 \dots$ (2)

Now equation of common chord of the circles is $S_{1} - S_{2} = 0 \Rightarrow 2 (g - g^{'}) x + 2 (f - f^{'}) y + c - c^{'} = 0 \dots$ (3)

Since cirlce (1) bisects the circumference of the circle (2), therefore, common chord will be the diameter of the circle (2) and hence center will be $(- g^{'}, - f^{'})$ of circle (2) will lie on the line (3)

$\Rightarrow - 2 (g - g^{'}) g^{'} - 2 (f - f^{'}) f^{'} + c - c^{'} = 0 \Rightarrow 2 g^{'} (g - g^{'}) + 2 f^{'} (f - f^{'}) = c - c^{'}$ .

The given circles are $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ and $x^{2} + y^{2} - 10 x - 12 y + 40 = 0$ .

The family of circles passing through their points of intersection is $S_{1} + λ S_{2} = 0$ .

So the required circle is $(x^{2} + y^{2} - 2 x - 4 y - 4) + λ (x^{2} + y^{2} - 10 x - 12 y + 40) = 0$ .

This simplifies to $(1 + λ) (x^{2} + y^{2}) + (- 2 - 10 λ) x + (- 4 - 12 λ) y + (- 4 + 40 λ) = 0$ .

$\Rightarrow x^{2} + y^{2} + \frac{- 2 - 10 λ}{1 + λ} x + \frac{- 4 - 12 λ}{1 + λ} y + \frac{- 4 + 40 λ}{1 + λ} = 0$ .

Comparing with $x^{2} + y^{2} + g x + f y + c = 0$ .

So $g = \frac{- 2 - 10 λ}{1 + λ}, f = \frac{- 4 - 12 λ}{1 + λ}$ , and $c = \frac{- 4 + 40 λ}{1 + λ}$ .

The radius condition is $g^{2} + f^{2} - c = 16$ . After solving we get $λ = 1$ .

Then the equation becomes $2 (x^{2} + y^{2}) - 12 x - 16 y + 36 = 0 \Rightarrow x^{2} + y^{2} - 6 x - 8 y + 18 = 0$ .

The given circles are $x^{2} + y^{2} - 6 x - 4 y + 9 = 0$ and $x^{2} + y^{2} - 8 x - 6 y + 23 = 0$ .

The common chord is obtained by subtracting the two equations.

So we get $(x^{2} + y^{2} - 6 x - 4 y + 9) - (x^{2} + y^{2} - 8 x - 6 y + 23) = 0$ .

This simplifies to $2 x + 2 y - 14 = 0$ or $x + y - 7 = 0$ .

From $x^{2} + y^{2} - 8 x - 6 y + 23 = 0$ the center is $(4, 3)$ .

Substitute $(4, 3)$ into $x + y - 7 = 0$ . We get $4 + 3 - 7 = 0$ .

So the common chord passes through the center of the second circle.

The radius of the second circle is $r^{2} = 16 + 9 - 23 = 2$ so $r = \sqrt{2}$ .

The perpendicular distance from the center $(4, 3)$ to the chord $x + y - 7 = 0$ is $\frac{| 4 + 3 - 7 |}{\sqrt{2}} = 0$ .

Hence, the chord passes through the center, so it is a diameter.

Therefore, the length of the chord is $2 r = 2 \sqrt{2}$ .

The given circles are $x^{2} + y^{2} + 2 x + 3 y + 1 = 0$ and $x^{2} + y^{2} + 4 x + 3 y + 2 = 0$ .

The common chord is obtained by subtracting the equations.

So we get $(x^{2} + y^{2} + 2 x + 3 y + 1) - (x^{2} + y^{2} + 4 x + 3 y + 2) = 0 \Rightarrow - 2 x - 1 = 0$ or $x = - \frac{1}{2}$ .

Now for a circle with diameter along a line, we use the fact that its center lies on the perpendicular bisector of the chord.

The midpoint of the chord lies on the line joining the centers of the two given circles.

The centers are $(- 1, - \frac{3}{2})$ and $(- 2, - \frac{3}{2})$ . So the line joining centers is $y = - \frac{3}{2}$ .

The midpoint of the chord is intersection of $x = - \frac{1}{2}$ and $y = - \frac{3}{2}$ .

So the center is $(- \frac{1}{2}, - \frac{3}{2})$ . Substitute $x = - \frac{1}{2}$ in first circle.

Then $\frac{1}{4} + y^{2} - 1 + 3 y + 1 = 0$ . This gives $y^{2} + 3 y + \frac{1}{4} = 0$ .

Solve to get $y = \frac{- 3 \pm 2 \sqrt{2}}{2}$ .

So the radius squared is $r^{2} = {(\sqrt{2})}^{2} = 2$ .

Hence, the equation of the circle is ${(x + \frac{1}{2})}^{2} + {(y + \frac{3}{2})}^{2} = 2$ .

The given circle is $x^{2} + y^{2} - 2 a x = 0$ and the chord is $y = m x$ .

Substitute $y = m x$ into the circle.

This gives $x^{2} + m^{2} x^{2} - 2 a x = 0 \Rightarrow (1 + m^{2}) x^{2} - 2 a x = 0$ .

So the points of intersection are $x = 0$ and $x = \frac{2 a}{1 + m^{2}}$ .

Thus, the points are $(0, 0)$ and $(\frac{2 a}{1 + m^{2}}, \frac{2 a m}{1 + m^{2}})$ .

These are the endpoints of the chord.

The equation of the circle with this chord as diameter is $(\frac{2 a}{1 + m^{2}}, \frac{2 a m}{1 + m^{2}})$ .

This gives $x (x - \frac{2 a}{1 + m^{2}}) + y (y - \frac{2 a m}{1 + m^{2}}) = 0 \Rightarrow (1 + m^{2}) (x^{2} + y^{2}) - 2 a (x + m y) = 0$ .

The given circles are $x^{2} + y^{2} - 6 x + 2 y + 4 = 0$ and $x^{2} + y^{2} + 2 x - 4 y - 6 = 0$ .

The family of circles passing through their points of intersection is

$S_{1} + λ S_{2} = 0$ . So the required circle is

$(x^{2} + y^{2} - 6 x + 2 y + 4) + λ (x^{2} + y^{2} + 2 x - 4 y - 6) = 0$ .

$\Rightarrow (1 + λ) (x^{2} + y^{2}) + (- 6 + 2 λ) x + (2 - 4 λ) y + (4 - 6 λ) = 0$ .

$\Rightarrow x^{2} + y^{2} + \frac{- 6 + 2 λ}{1 + λ} x + \frac{2 - 4 λ}{1 + λ} y + \frac{4 - 6 λ}{1 + λ} = 0$ .

The center is $(\frac{6 - 2 λ}{2 (1 + λ)}, \frac{- 2 + 4 λ}{2 (1 + λ)})$ .

Since the center lies on $y = x$ , equate the coordinates. So $6 - 2 λ = - 2 + 4 λ$ .

This gives $8 = 6 λ$ so $λ = \frac{4}{3}$ .

Then $1 + λ = \frac{7}{3}$ . So the equation becomes

$\frac{7}{3} (x^{2} + y^{2}) + (- \frac{10}{3}) x + (- \frac{10}{3}) y - 4 = 0 \Rightarrow 7 (x^{2} + y^{2}) - 10 x - 10 y - 12 = 0$ .

The given equation is $x^{2} + y^{2} + 2 (3 + p) x + 2 (3 - p) y + 4 = 0$ .

This is of the form $x^{2} + y^{2} + g x + f y + c = 0$ so it represents a circle for all values of $p$ .

$\Rightarrow x^{2} + y^{2} + 6 x + 6 y + 4 + 2 p (x - y) = 0$ .

For fixed points, the equation must be satisfied for all values of $p$ .

So the coefficient of $p$ must be zero and the remaining part must also be zero.

Thus, we get $x - y = 0$ and $x^{2} + y^{2} + 6 x + 6 y + 4 = 0$ .

From $x - y = 0$ we get $y = x$ .

Substitute into the second equation $x^{2} + x^{2} + 6 x + 6 x + 4 = 0 \Rightarrow 2 x^{2} + 12 x + 4 = 0$ .

Solving gives $x = - 3 \pm \sqrt{7}$ .

Since $y = x$ , the fixed points are $(- 3 + \sqrt{7}, - 3 + \sqrt{7})$ and $(- 3 - \sqrt{7}, - 3 - \sqrt{7})$ .

The given circles are $x^{2} + y^{2} - 4 a^{2} = 0$ and $x^{2} + y^{2} - 2 x - 4 y + 4 = 0$ .

The family of circles through their intersection is $(x^{2} + y^{2} - 4 a^{2}) + λ (x^{2} + y^{2} - 2 x - 4 y + 4) = 0$ .

This gives $(1 + λ) (x^{2} + y^{2}) - 2 λ x - 4 λ y + (- 4 a^{2} + 4 λ) = 0$ .

The center is $(\frac{λ}{1 + λ}, \frac{2 λ}{1 + λ})$ .

Since the circle touches $x + 2 y = 0$ , the distance from center equals radius.

This gives $\frac{4 a^{2} - 4 λ}{1 + λ} = 0$ so $λ = a^{2}$ .

Hence, the required circle is $(1 + a^{2}) (x^{2} + y^{2}) - 2 a^{2} x - 4 a^{2} y + 4 a^{2} (1 - a^{2}) = 0$ .

The given circle is $x^{2} + y^{2} - x - y = 0$ and the line is $x + y = 1$ .

The family of circles passing through their intersection points is $x^{2} + y^{2} - x - y + λ (x + y - 1) = 0$ .

This gives $x^{2} + y^{2} + (- 1 + λ) x + (- 1 + λ) y - λ = 0$ .

Since the circle passes through $(1, 1)$ , substitute it.

So $1 + 1 + (- 1 + λ) + (- 1 + λ) - λ = 0 \Rightarrow λ = - 2$ .

Hence, the required circle is $x^{2} + y^{2} - 3 x - 3 y + 2 = 0$ .

The given circle is $x^{2} + y^{2} = a^{2}$ and the line is $p x + q y - 1 = 0$ .

Let $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ be the endpoints of this chord.

Equation of the circle is $x^{2} + y^{2} - (x_{1} + x_{2}) x - (y_{1} + y_{2}) y + (x_{1} x_{2} + y_{1} y_{2}) = 0$ .

Now $(x_{1} + x_{2}, y_{1} + y_{2})$ is twice the midpoint of the chord.

The midpoint is the foot of the perpendicular from the center $(0, 0)$ to the line.

So midpoint is $(\frac{p}{p^{2} + q^{2}}, \frac{q}{p^{2} + q^{2}})$ .

Hence, $x_{1} + x_{2} = \frac{2 p}{p^{2} + q^{2}}$ and $y_{1} + y_{2} = \frac{2 q}{p^{2} + q^{2}}$ .

Also both points satisfy $p x + q y = 1$ . So $p (x_{1} + x_{2}) + q (y_{1} + y_{2}) = 2$ .

Thus, $x_{1} x_{2} + y_{1} y_{2} = \frac{1}{p^{2} + q^{2}}$ . Substitute in the diameter form.

Hence, the required circle is $x^{2} + y^{2} - \frac{2 p}{p^{2} + q^{2}} x - \frac{2 q}{p^{2} + q^{2}} y + \frac{1}{p^{2} + q^{2}} = 0$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and the external point is $A (α, β)$ .

Let $P$ and $Q$ be the points of contact of tangents from $A$ . The chord of contact of $A$ with respect to the circle is $T = 0$ .

So the equation of chord $P Q$ is $x α + y β + g (x + α) + f (y + β) + c = 0$ .

The circumcircle of $△ P Q R$ where $R$ is the center of the given circle is obtained by combining $S = 0$ and $T = 0$ .

So its equation is $S + λ T = 0$ .

Since it passes through $A (α, β)$ , we substitute it. Then $S_{1} + λ T_{1} = 0$ .

Here $T_{1} = S_{1}$ . So we get $S_{1} (1 + λ) = 0$ which gives $λ = - 1$ .

Hence, the required circle is $S - T = 0$ .

So the equation is $x^{2} + y^{2} + 2 g x + 2 f y + c - [x α + y β + g (x + α) + f (y + β) + c] = 0$ .

$\Rightarrow x^{2} + y^{2} + (g - α) x + (f - β) y - (g α + f β) = 0$ .

Substituting the value of $y = \frac{1}{4} (3 x - c)$ in the equation of the circle gives us

$x^{2} + \frac{1}{16} (3 x - c)^{2} - 4 x - 8. \frac{1}{4} (3 x - c) - 5 = 0 \Rightarrow 25 x^{2} - 2 (80 + 3 c) x + c^{2} + 32 c - 80 = 0$

The given line and circle will intersect if the above quadratic equation's roots are real i.e. discrimininate $> 0$

$\Rightarrow 4 (80 + 3 c)^{2} - 100 (c^{2} + 32 c - 80) > 0 => c^{2} + 20 c - 525 < 0 \Rightarrow - 35 < c < 15$ .

Center of the circle is $(- \frac{3}{2}, \frac{1}{2})$ and radius is $\frac{5}{\sqrt{2}}$ . Let $l$ be the length of the perpendicular from the center to the given line then

$l = \frac{| 4 (- \frac{3}{2}) - \frac{3}{2} - 5 |}{5} = \frac{5}{2}$

Hence, length of the chord is $2 \sqrt{\frac{25}{2} - \frac{25}{4}} = \frac{5}{2}$ .

Center of the circle is $(0, 0)$ and its perpendicular distance from the line is $\frac{| a \sqrt{2} |}{\sqrt{2}} = a$ , which is equal to the radius of the circle.

Hence, the given circle touches the given line. Let $(α, β)$ be the point of contact. Then equation of tangent is given by $α x + β y - a^{2} = 0$

Comparing the coefficients with the given equation of the line we have

$\frac{α}{1} = \frac{β}{- 1} = - \frac{a^{2}}{a \sqrt{2}} \Rightarrow (α, β) = (- \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}})$ .

Center of the given circle is $(3, - 2)$ and radius is $5$ . Equation of any line parallel to given line is $4 x + 3 y + k = 0$ .

Since this point is tangent to the given circle, therefore,

$\frac{| 4.3 - 3.2 + k |}{5} = 5 \Rightarrow | 6 + k | = 25 \Rightarrow k = 19, - 31$ .

Given circles are $S_{1} = x^{2} + y^{2} - 6 x = 0 \dots$ (1) and $S_{2} = x^{2} + y^{2} + 2 x = 0 \dots$ (2).

Figure 11.7.

Let $A$ and $B$ are the centers and $r_{1}$ and $r_{2}$ the radii of $S_{1}$ and $S_{2}$ respectively.

$A = (3, 0), B = (- 1, 0), r_{1} = 3, r_{2} = 1 ∴ r_{1} + r_{2} = 4$ . Hence two circles touch each other. Thus, there will be three common tangents.

Equation of chord is given by $S_{1} - S_{2} = x = 0$ , when $x = 0, y = 0$ . Thus, $x = 0$ is a common tangent.

Let $y = m x + c$ be a common tangent to the given circles then

$f r a c | 3 m + c | \sqrt{1 + m^{2}} = 3$ and $\frac{| - m + c |}{\sqrt{1 + m^{2}}} = 1$

Solving these two equations gives us $c = \pm \sqrt{3}$ and $m = \pm \frac{1}{\sqrt{3}}$

Thus, common tangents are $x = 0, y = \frac{x}{\sqrt{3}} + s q r t (3)$ , and $y = - \frac{x}{\sqrt{3}} - \sqrt{3}$

Let $P, Q, R$ be points of intersections of these three lines then $P = (0, \sqrt{3}), Q = (- 3, 0)$ and $R = (0, - \sqrt{3})$ .

It is trivial to prove that $△ P Q R$ is an equilateral triangle.

Given that the biggest circle is $x^{2} + y^{2} = 1 \dots$ (1). Since the radiio of the circles are in A.P. let the commond difference be $d$ .

Thus, other two circles will be $x^{2} + y^{2} = (1 - d)^{2} \dots$ (2) and $x^{2} + y^{2} = (1 - 2 d)^{2} \dots$ (3)

Given line is $y = x + 1$ . Putting in (1) gives us $x^{2} + (x + 1)^{2} = 1 \Rightarrow x = 0, - 1$

Similarly with (2) we have $x^{2} + (x + 1)^{2} = (1 - d^{2}) \Rightarrow 2 x^{2} + 2 x + 2 d - d^{2} = 0$

Since the points are real and distinct, therefore, $4 - 8 (2 d - d^{2}) > 0 \Rightarrow 2 d^{2} - 4 d + 1 > 0$

$\Rightarrow 1 - \frac{1}{\sqrt{2}} > d > 1 + \frac{1}{\sqrt{2}}$

Similarly with (3) we have $x^{2} + (x + 1)^{2} = (1 - 2 d)^{2}$ . Proceeding similarly we obtain

$\frac{2 - \sqrt{2}}{4} > d > \frac{2 + \sqrt{2}}{4}$ . However, $d < 1$ .

Thus, we have $0 < d < \frac{2 - \sqrt{2}}{4}$ .

Given $4 l^{2} - 5 m^{2} + 6 l + 1 = 0 \dots$ (1) and line is $l x + m y + 1 = 0 \dots$ (2)

Let the center of the circle be $(α, β)$ with radius $a$ . Then

$\frac{| l a l p h a + m b e t a + 1 |}{\sqrt{l^{2} + m^{2}}} = a$

$\Rightarrow l^{2} α^{2} + m^{2} β^{2} + 1 + 2 l m α β + 2 l α + 2 m β = a^{2} l^{2} + a^{2} m^{2}$

$=> {(α^{2} - a^{2})}^{2} + (β^{2} - a^{2}) m^{2} + 2 l m α β + 2 α l + 2 m β + 1 = 0$

Comparing this with (1)

$α^{2} - a^{2} = 4, β^{2} - a^{2} = - 5, α = 3, β = 0$ . Thus, $a = \sqrt{5}$ .

Hence, the circle has center $(3, 0)$ and radius $\sqrt{5}$ .

Given cicle is $x^{2} + y^{2} - 4 x - 6 y + 9 = 0 d o t s$ (1). Its center is $C (2, 3)$ and its radius is $2$ .

Figure 11.8.

Let $O P$ be a tangent and let $y$ -axis(which is a tanegent) touch the circle at $N$ . Then $∠ P O X$ will be minimum when $O P$ is tangent to the circle.

Let $∠ P O X = θ$ then $∠ L C P = θ$

Now $C P = 2, O C = \sqrt{2^{2} + 3^{2}} = \sqrt{13}$

$O P = \sqrt{O C^{2} - C P^{2}} = 3$

From figure $O M = O L + L M = O L + H P \Rightarrow O P \cos θ = 2 + 2 \sin θ$ or $3 \cos θ = 2 + 2 \sin θ$

$\Rightarrow \cos θ = \frac{12}{13}, \sin θ = \frac{5}{13}$

$P = (\frac{36}{13}, \frac{15}{13})$

$O P$ will be maximum if $P$ becomes the point where extended part of $O C$ cuts the circle. Let this point be $P_{2}$ .

$O P_{2} = O C + r = \sqrt{13} + 2$

Slope is $\frac{3}{2} = \tan α$ (let) $\Rightarrow P_{2} = (2 + \frac{4}{\sqrt{13}}, 3 + \frac{6}{\sqrt{13}})$ .

Given circle is $x^{2} + y^{2} - 2 a x - 2 a y + a^{2} = 0$ . First we find its point of contact with $x$ -axis i.e. $y = 0$ .

Putting $y = 0, x^{2} - 2 a x + a^{2} = 0 \Rightarrow x = a$ . Thus, point of contact is $(a, 0)$ .

Then we put $x = 0$ to get $y = a$ . Thus, point of contact is $(0, a)$ (because we get only one point in both the cases the circle touches the axes.)

The given circle is $x^{2} + y^{2} - 16 = 0$ so its center is $(0, 0)$ and radius is $4$ .

The given points are $(2, 3)$ and $(1, 2)$ . The midpoint is $(\frac{3}{2}, \frac{5}{2})$ .

The slope of the line joining the points is $1$ so the perpendicular slope is $- 1$ .

Hence the chord is the line through $(\frac{3}{2}, \frac{5}{2})$ with slope $- 1$ .

So its equation is $y - \frac{5}{2} = - 1 (x - \frac{3}{2}) \Rightarrow x + y - 4 = 0$ .

So $d = \frac{| 0 + 0 - 4 |}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$ .

The length of the chord is $2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{16 - 8} = 4 \sqrt{2}$ .

The given circle is $x^{2} + y^{2} - 14 x + 4 y + 28 = 0 \Rightarrow (x - 7)^{2} + (y + 2)^{2} = 25$ .

So the center is $(7, - 2)$ and radius is $5$ .

The given line is $x - 7 y + 4 = 0$ . The perpendicular distance from the center to the line is

$\frac{| 7 - 7 (- 2) + 4 |}{\sqrt{1 + 49}} = \frac{5}{\sqrt{2}}$ .

The length of the chord is $2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{25 - \frac{25}{2}} = 5 \sqrt{2}$ .

The midpoint of the chord is the foot of the perpendicular from the center to the line.

Using formula, midpoint is $(7 - 1 \times \frac{25}{50}, - 2 + 7 \times \frac{25}{50}) \Rightarrow (\frac{13}{2}, \frac{3}{2})$ .

The given circles are $x^{2} + y^{2} + 3 x + 5 y + 4 = 0$ and $x^{2} + y^{2} + 5 x + 3 y + 4 = 0$ .

So we get $- 2 x + 2 y = 0$ or $y = x$ .

First circle is ${(x + \frac{3}{2})}^{2} + {(y + \frac{5}{2})}^{2} = \frac{9}{2}$ . So the center is $(- \frac{3}{2}, - \frac{5}{2})$ and radius is $\frac{3}{\sqrt{2}}$ .

The distance from the center to the line $y = x$ is $\frac{| - 3 / 2 + 5 / 2 |}{\sqrt{2}} = \frac{1}{\sqrt{2}}$ .

The length of the common chord is $2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{\frac{9}{2} - \frac{1}{2}} = 4$ .

Hence, the length of the common chord is $4$ .

The given circles are $x^{2} + y^{2} + 2 x + 3 y + 1 = 0$ and $x^{2} + y^{2} + 4 x + 3 y + 2 = 0$ .

So we get $- 2 x - 1 = 0$ or $x = - 1 / 2$ upon solving.

First circle is $(x + 1)^{2} + {(y + \frac{3}{2})}^{2} = \frac{9}{4}$ . So the center is $(- 1, - \frac{3}{2})$ and radius is $\frac{3}{2}$ .

The distance from the center to the chord $x = - \frac{1}{2}$ is $\frac{| - 1 + 1 / 2 |}{1} = \frac{1}{2}$ .

The length of the chord is $2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{\frac{9}{4} - \frac{1}{4}} = 2 \sqrt{2}$ .

The given circles are $(x - a)^{2} + (y - b)^{2} = c^{2}$ and $(x - b)^{2} + (y - a)^{2} = c^{2}$ . So $(x - a)^{2} + (y - b)^{2} - (x - b)^{2} - (y - a)^{2} = 0 \Rightarrow 2 (a - b) (y - x) = 0$ so $y = x$ .

So the common chord lies on $y = x$ .

The center is $(a, b)$ and radius is $c$ of the first circle.

The perpendicular distance from $(a, b)$ to the line $y = x$ is $\frac{| a - b |}{\sqrt{2}}$ .

The length of the chord is $2 \sqrt{c^{2} - d^{2}}$ . So it is $2 \sqrt{c^{2} - \frac{(a - b)^{2}}{2}}$ .

This simplifies to $\sqrt{4 c^{2} - 2 (a - b)^{2}}$ . Hencem the length of the common chord is $\sqrt{4 c^{2} - 2 (a - b)^{2}}$ .

For the circles to touch, the chord length must be zero. So $4 c^{2} - 2 (a - b)^{2} = 0$ . Hence, the condition is $2 c^{2} = (a - b)^{2}$ .

The given circles are $x^{2} + y^{2} + 2 h x + a^{2} = 0$ and $x^{2} + y^{2} - 2 k y - a^{2} = 0$ .

Common chord's equaiton is $2 h x + a^{2} + 2 k y + a^{2} = 0 \Rightarrow h x + k y + a^{2} = 0$ .

First circle's center is $(- h, 0)$ and radius squared is $h^{2} - a^{2}$ .

The perpendicular distance from the center to the chord is $\frac{| - h^{2} + a^{2} |}{\sqrt{h^{2} + k^{2}}}$ .

So $d^{2} = \frac{{(h^{2} - a^{2})}^{2}}{h^{2} + k^{2}}$ .

The length of the chord is $2 \sqrt{r^{2} - d^{2}} \Rightarrow 2 \sqrt{(h^{2} - a^{2}) - \frac{{(h^{2} - a^{2})}^{2}}{h^{2} + k^{2}}}$ .

$= 2 \sqrt{(h^{2} - a^{2}) (1 - \frac{h^{2} - a^{2}}{h^{2} + k^{2}})}$ $= 2 \sqrt{\frac{(h^{2} - a^{2}) (k^{2} + a^{2})}{h^{2} + k^{2}}}$ .

The given circles are $x^{2} + y^{2} + a x + b y + c = 0$ and $x^{2} + y^{2} + b x + a y + c = 0$ .

So chord is $(a - b) x + (b - a) y = 0$ which gives $x = y$ .

First circle's center is $(- \frac{a}{2}, - \frac{b}{2})$ and radius squared is $\frac{a^{2} + b^{2}}{4} - c$ .

The perpendicular distance from the center to the line $x - y = 0$ is $\frac{| - a / 2 + b / 2 |}{\sqrt{2}} = \frac{| a - b |}{2 \sqrt{2}}$ .

So $d^{2} = \frac{(a - b)^{2}}{8}$ .

The length of the chord is $2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{\frac{a^{2} + b^{2}}{4} - c - \frac{(a - b)^{2}}{8}} = 2 \sqrt{\frac{(a + b)^{2}}{8} - c}$ .

So the length is $\sqrt{\frac{(a + b)^{2}}{2} - 4 c}$ .

The center of the given circle is origin and radius is $a$ . The length of perpendicular from center to tangent is equal to radius. Therefore,

$\frac{r}{\sqrt{p^{2} + q^{2}}} = a \Rightarrow r^{2} = a^{2} (p^{2} + q^{2})$ .

Center of the given circle is $(3, - 5)$ and radius is $\sqrt{3^{2} + 5^{2} + 66} = 10$ .

Length of the perpendicular on the given line from center is $\frac{| 4.3 + 3.5 + 23 |}{5} = 10$ , which is equal to the radius of the circle.

Hence, the given circle touches the given line.

Center of the given circle is the origin and radius is $a$ .

Length of the perpendicular on the given line from center is $\frac{| - a |}{\sqrt{\sin^{2} θ + \cos^{2} θ}} = a$ , which is equaal to radius.

Thus, the given line touches the given circle.

Center of the given circle is the origin and radius is $a$ .

Length of the perpendicular on the given line from center is $\frac{| - 1 |}{\sqrt{l^{2} + m^{2}}} = a \Rightarrow l^{2} + m^{2} = a^{- 2}$ .

Thus, locus of $(l, m)$ is the circle $x^{2} + y^{2} = a^{- 2}$ .

Given circle has center $(2, 4)$ and radius $\sqrt{2^{2} + 4^{2} + 5} = 5$ .

For the given line to touch the circle length of perpendicular from center to the line must be equal to the radius of the circle. Thus,

$\frac{| 3.2 - 4.4 - λ |}{5} = 5 \Rightarrow 10 + λ = \pm 25 \Rightarrow λ = 15, - 35$ .

The given line is $(x - 1) \cos θ + (y - 1) \sin θ = 1$ .

Expand to get $x \cos θ + y \sin θ - \cos θ - \sin θ - 1 = 0$ .

For a fixed point $(h, k)$ to be the center of a circle touched by all these lines, the perpendicular distance from $(h, k)$ to the line must be constant.

So distance is $| h \cos θ + k \sin θ - \cos θ - \sin θ - 1 |$ .

This becomes $| (h - 1) \cos θ + (k - 1) \sin θ - 1 |$ .

For this to be independent of $θ$ , we must have $h - 1 = 0$ and $k - 1 = 0$ .

So the center is $(1, 1)$ . Now the distance becomes constant equal to $1$ .

Hence, the radius is $1$ .

Therefore, the required circle is $(x - 1)^{2} + (y - 1)^{2} = 1$ .

The given line is $3 x - 16 y = 10$ . So the required tangents are of the form $3 x - 16 y + c = 0$ .

For the circle $x^{2} + y^{2} = 16$ , the center is $(0, 0)$ and radius is $4$ .

The distance from the center to the tangent must be equal to the radius.

So $\frac{| c |}{\sqrt{3^{2} + (- 16)^{2}}} = 4 \Rightarrow | c | = 4 \sqrt{265}$ . Hence, $c = \pm 4 \sqrt{265}$ .

Therefore the required tangents are $3 x - 16 y + 4 \sqrt{265} = 0$ and $3 x - 16 y - 4 \sqrt{265} = 0$ .

The given circle is $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ . So the center is $(1, 2)$ and radius is $3$ .

First consider tangents parallel to $3 x - 4 y - 1 = 0$ . Such lines are of the form $3 x - 4 y + c = 0$ .

The distance from the center to the line equals the radius. So $\frac{| 3 (1) - 4 (2) + c |}{5} = 3$ .

$\Rightarrow | - 5 + c | = 15 \Rightarrow c = 20$ or $c = - 10$ .

Hence, the tangents are $3 x - 4 y + 20 = 0$ and $3 x - 4 y - 10 = 0$ .

Now consider tangents perpendicular to $3 x - 4 y - 1 = 0$ .

Slope of the given line is $\frac{3}{4}$ so perpendicular slope is $- \frac{4}{3}$ . So the tangents are of the form $4 x + 3 y + c = 0$ .

Again distance condition gives $\frac{| 4 (1) + 3 (2) + c |}{5} = 3 \Rightarrow c = 5$ or $c = - 25$ .

Hence, the tangents are $4 x + 3 y + 5 = 0$ and $4 x + 3 y - 25 = 0$ .

The given circle is $x^{2} + y^{2} - 5 x + 5 y = 0$ .

So the center is $(\frac{5}{2}, - \frac{5}{2})$ and radius is $\frac{5}{\sqrt{2}}$ .

The given line is $7 y - x - 5 = 0$ .

The distance from the center to this line is $\frac{| - \frac{5}{2} - \frac{35}{2} - 5 |}{\sqrt{50}} = \frac{5}{\sqrt{2}}$ .

So the line touches the circle. Now the other parallel tangent is of the form $7 y - x + c = 0$ .

Again use the distance condition. So $\frac{| - \frac{5}{2} - \frac{35}{2} + c |}{\sqrt{50}} = \frac{5}{\sqrt{2}}$ .

This gives $| c - 20 | = 25$ . So $c = 45$ or $c = - 5$ .

Since $c = - 5$ gives the given line, the other tangent is $7 y - x + 45 = 0$ .

The given circle is $x^{2} + y^{2} = 15$ so the center is $(0, 0)$ and radius is $\sqrt{15}$ .

The given line $4 x - y + 6 = 0$ has slope $4$ . So the required tangents have slope $- \frac{1}{4}$ .

Hence, their equations are of the form $x + 4 y + c = 0$ .

The distance from the center to the tangent must be equal to the radius.

So $\frac{| c |}{\sqrt{1 + 16}} = \sqrt{15}$ . Thus $\frac{| c |}{\sqrt{17}} = \sqrt{15}$ .

$\Rightarrow | c | = \sqrt{255}$ . Hence, $c = \pm s q r t 255$ .

Therefore the required tangents are $x + 4 y + \sqrt{255} = 0$ and $x + 4 y - \sqrt{255} = 0$ .

The given circle is $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ . So the center is $(3, - 2)$ and radius is $4$ .

The given line $y = 2 x - 1$ has slope $2$ . So the required tangents have slope $- \frac{1}{2}$ .

Hence, their equations are of the form $x + 2 y + c = 0$ .

The distance from the center to the tangent equals the radius. So $\frac{| 3 + 2 (- 2) + c |}{\sqrt{5}} = 4$ .

So $c = 1 \pm 4 \sqrt{5}$ .

Hence, the required tangents are $x + 2 y + 1 + 4 \sqrt{5} = 0$ and $x + 2 y + 1 - 4 \sqrt{5} = 0$ .

The given circle is $x^{2} + y^{2} = 25$ so the center is $(0, 0)$ and radius is $5$ .

A line making an angle $60^{\circ}$ with the positive $x$ -axis has slope $\tan 60^{\circ} = \sqrt{3}$ .

So the required tangents are of the form $y = \sqrt{3} x + c$ .

The distance from the center to the tangent must be equal to the radius. So $\frac{| c |}{\sqrt{1 + 3}} = 5 \Rightarrow | c | = 10$ . Hence, $c = \pm 10$ .

Therefore, the required tangents are $y = \sqrt{3} x + 10$ and $y = \sqrt{3} x - 10$ .

The given pair of lines is $x^{2} - y^{2} + 2 y - 1 = 0$ .

Rewrite it as $x^{2} - (y - 1)^{2} = 0$ . So the lines are $x = y - 1$ and $x = 1 - y$ .

The family of circles touching both lines has its center on the angle bisectors.

The angle bisectors are $x = 0$ and $y = 1$ . First take center $(0, k)$ .

The radius is the distance from $(0, k)$ to either line. So $r = \frac{| 0 - k + 1 |}{\sqrt{2}}$ .

Hence, the circle is $x^{2} + (y - k)^{2} = \frac{(k - 1)^{2}}{2}$ .

This gives one family. Now take center $(h, 1)$ . The radius is $\frac{| h |}{\sqrt{2}}$ .

Hence, the circle is $(x - h)^{2} + (y - 1)^{2} = \frac{h^{2}}{2}$ .

Let $A$ and $B$ be the centers and $r_{1}$ and $r_{2}$ the radii of the given circles respectively. Thus, $A = (1, 2), B = (0, 4), r_{1} = \sqrt{5}$ , and $r_{2} = 2 \sqrt{5}$ .

$A B = \sqrt{(1 - 0)^{2} + (2 - 4)^{2}} = \sqrt{5}$

$r_{1} + r_{2} = 3 \sqrt{5}$ and $| r_{1} - r_{2} | = \sqrt{5}$

Thus, $A B = | r_{1} - r_{2} |$ , hence, the two circles touch each other internally.

The centers of the given circles are $A (- a, 0)$ and $B (0, - b)$ and radii are $r_{1} = \sqrt{a^{2} - c^{2}}$ and $r_{2} = \sqrt{b^{2} - c^{2}}$ respectively.

The circle will touch internally or externally if $A B = r_{1} + r_{2}$ or $A B = | r_{1} - r_{2} |$

$A B^{2} = (r_{1} \pm r_{2})^{2} \Rightarrow a^{2} + b^{2} = r_{1}^{2} + r_{2}^{2} \pm 2 r_{1} r_{2}$

Substituting the values and squaring we get $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}$ .

The centers of the given circles are $A (0, 0)$ and $B (2 a, 0)$ respectively, and radii are $a$ for both.

Distance between centers $A B = 2 a = r_{1} + r_{2}$ . Hence, the circles touch each other externally.

Let the equations of the circles touching both the given circles be $(x - α)^{2} + (y - β)^{2} = a^{2}$ with center $C = (α, β)$ and radius $a$ .

$A C = r_{1} + r_{3} = 2 a \Rightarrow α^{2} + β^{2} = 4 a^{2}$ , and similarly, $B C = r_{2} + r_{3} \Rightarrow (2 α - a)^{2} + β^{2} = 4 a^{2}$

$\Rightarrow α = a, β = \pm \sqrt{3} a$ , and thus, we have our required circles.

The given circles are $x^{2} + y^{2} + 2 x + 2 y + 1 = 0$ and $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ .

First circle is $(x + 1)^{2} + (y + 1)^{2} = 1$ . Hence, the center is $(- 1, - 1)$ and radius is $1$ .

Second circle is $(x - 2)^{2} + (y - 3)^{2} = 16$ . Hencem the center is $(2, 3)$ and radius is $4$ .

Now find the distance between the centers. So $d = \sqrt{(2 + 1)^{2} + (3 + 1)^{2}} = 5$ .

Also $r_{1} + r_{2} = 1 + 4 = 5$ . Since $d = r_{1} + r_{2}$ , the circles touch externally.

The given circles have centers $(a, b)$ and $(b, a)$ and both have radius $c$ .

The distance between the centers is $\sqrt{(a - b)^{2} + (b - a)^{2}} = \sqrt{2 (a - b)^{2}} = \sqrt{2} | a - b |$ .

For the circles to touch externally, the distance must be equal to $2 c$ .

So $\sqrt{2} | a - b | = 2 c$ . Hence $| a - b | = \sqrt{2} c$ .

Thus, $a - b = \pm \sqrt{2} c$ . So the condition is $a = b \pm \sqrt{2} c$ .

The given circles are $x^{2} + y^{2} + 2 u x + 2 v y = 0$ and $x^{2} + y^{2} + 2 u_{1} x + 2 v_{1} y = 0$ .

Their centers are $(- u, - v)$ and $(- u_{1}, - v_{1})$ . Their radii are $\sqrt{u^{2} + v^{2}}$ and $\sqrt{u_{1}^{2} + v_{1}^{2}}$ .

The distance between the centers is $\sqrt{(u - u_{1})^{2} + (v - v_{1})^{2}}$ .

For the circles to touch, we must have $\sqrt{(u - u_{1})^{2} + (v - v_{1})^{2}} = \sqrt{u^{2} + v^{2}} \pm \sqrt{u_{1}^{2} + v_{1}^{2}}$ .

$\Rightarrow (u - u_{1})^{2} + (v - v_{1})^{2} = u^{2} + v^{2} + u_{1}^{2} + v_{1}^{2} \pm 2 \sqrt{(u^{2} + v^{2}) (u_{1}^{2} + v_{1}^{2})}$ .

$\Rightarrow u^{2} + v^{2} + u_{1}^{2} + v_{1}^{2} - 2 (u u_{1} + v v_{1})$ equals the right side.

$\Rightarrow - 2 (u u_{1} + v v_{1}) = \pm 2 \sqrt{(u^{2} + v^{2}) (u_{1}^{2} + v_{1}^{2})}$ .

$\Rightarrow u u_{1} + v v_{1} = \mp \sqrt{(u^{2} + v^{2}) (u_{1}^{2} + v_{1}^{2})}$ .

$\Rightarrow (u u_{1} + v v_{1})^{2} = (u^{2} + v^{2}) (u_{1}^{2} + v_{1}^{2})$ .

$\Rightarrow u^{2} u_{1}^{2} + v^{2} v_{1}^{2} + 2 u u_{1} v v_{1} = u^{2} u_{1}^{2} + u^{2} v_{1}^{2} + v^{2} u_{1}^{2} + v^{2} v_{1}^{2}$ .

$\Rightarrow 2 u u_{1} v v_{1} = u^{2} v_{1}^{2} + v^{2} u_{1}^{2} \Rightarrow u^{2} v_{1}^{2} - 2 u u_{1} v v_{1} + v^{2} u_{1}^{2} = 0$ .

So $(u v_{1} - u_{1} v)^{2} = 0$ . Hence, $u v_{1} = u_{1} v$ .

Given circle is $x^{2} + y^{2} = 2^{2} \dots$ (1)

Figure 11.9.

For point $P (1, \frac{1}{2}), x^{2} + y^{2} - 4 = 1 + 1 / 4 - 4 < 0$ , hence, the point lies inside the circle.

Let $A B$ be any chord of the circle through $P$ . Let $O L ⊥ A B$ , then $L$ , the will be the middle point of $A B$ .

$A B = 2 A L = \sqrt{O A^{2} - O L^{2}} = 2 \sqrt{4 - O P^{2} + L P^{2}}$

Since $P$ and $O$ are fixed points $O P$ is fixed.

$A B$ will be minimum if $L P$ is minimum and minimum value of $L P$ is $0$ , when $P$ coincides with $L$ .

Thus, minimum value of $A B = 2 \sqrt{4 - (1 + \frac{1}{4})} = \sqrt{11}$ .

Let $A B$ be the chord whose equation is $x + y - 1 = 0 \dots$ (1) and given circle is $x^{2} + y^{2} - 4 y = 0 \dots$ (2).

Figure 11.10.

Center of the circle is $(0, 2)$ . Let $L$ be the mid-point of the chord. Let $∠ A C B = 2 θ$ then $∠ A C L = ∠ B C L = θ$

$C L =$ length of perpendicular from $C$ to line (1) $= \frac{| 0 + 2 - 1 |}{\sqrt{2}} = \frac{1}{\sqrt{2}}$

$A C = 2$ (radius of the circle).

From $△ A C L, \cos θ = \frac{C L}{A C} = \frac{1}{2 \sqrt{2}}$

Now angle at circumference $= \frac{1}{2} \times$ angle at the center $= θ = \cos^{- 1} \frac{1}{2 \sqrt{2}}$ .

Let $A P B$ be a semicircle and $A B$ be a diameter. Let $O$ be the middle point of $A B$ . We take $O$ as the origin and $O B$ as $x$ -axis. Let $r$ be the radius of the semi-circle. Then $O = (0, 0), A = (- r, 0)$ and $B = (r, 0)$ .

Let $P = (x, y)$ . Now $O P^{2} = r^{2} ∴ x^{2} + y^{2} = r^{2}$

and $A P^{2} + P B^{2} = [(x + a)^{2} + y^{2}] + [(x - a)^{2} + y^{2}] = 2 (x^{2} + y^{2} + r^{2}) = 4 r^{2} = (2 a)^{2} = A B^{2}$

$∴ ∠ A P B = 90^{\circ}$ .

Figure 11.11.

Given $A B = 13$ m, $A C = 5$ m. Let $O L ⊥ A C$ , then $L$ is the mid-point of $A C$ .

Figure 11.12.

$A L = 2.5$ m and $A O = 6.5$ m. From $△ A L O, \cos θ = \frac{A L}{A O} = \frac{5}{13}$

Slope of $B C_{1} = \tan (90^{\circ} + θ) = - \cot θ = - \frac{5}{12}$

Slope of $B C_{2} = \tan (- 90^{\circ} - θ) = \frac{5}{12}$

Equations of $B C_{1}$ and $B C_{2}$ are $y = - \frac{5}{12} (x - \frac{13}{2})$ and $y = \frac{5}{12} (x - \frac{13}{2})$

The join equation is $100 x^{2} - 576 y^{2} - 1300 x + 4225 = 0$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and the internal point is $(α, β)$ .

Let the center be $(- g, - f)$ and radius be $r$ where $r^{2} = g^{2} + f^{2} - c$ .

Let a chord through $(a l p h a, b e t a)$ be at perpendicular distance $d$ from the center.

The length of the chord is $2 \sqrt{r^{2} - d^{2}}$ .

For a fixed point inside the circle, the least chord occurs when the chord is perpendicular to the line joining the center and the point.

In that case the distance from the center to the chord is the distance between the center and the point.

So $d^{2} = (α + g)^{2} + (β + f)^{2}$ .

Thus, the least length is $2 \sqrt{r^{2} - d^{2}} \Rightarrow 2 \sqrt{g^{2} + f^{2} - c - ((α + g)^{2} + (β + f)^{2})}$ .

$= 2 \sqrt{- (α^{2} + β^{2} + 2 g α + 2 f β + c)}$ .

Equation of any curve through the point of intersection of given lines and coordinate axes is $(a_{1} x + b_{1} y + c_{1}) (a_{2} x + b_{2} y + c_{2}) + k x y = 0$

If this is a circle then coeff. of $x^{2} =$ coeff. of $y^{2} \Rightarrow a_{1} a_{2} = b_{1} b_{2}$ .

The given circle is $x^{2} + y^{2} = k^{2}$ so the radius is $k$ . The line is $y - x - 3 = 0$ .

The perpendicular distance from the center $(0, 0)$ to the chord is

$d = \frac{| - 3 |}{\sqrt{2}} = \frac{3}{\sqrt{2}}$ .

Let the chord subtend angle $30^{\circ}$ at a point on the major segment.

Then the angle subtended at the center by the chord is $360^{\circ} - 2 \times 30^{\circ} = 300^{\circ}$ .

The half angle at the center is $150^{\circ}$ .

So $\cos 150^{\circ} = \frac{d}{l}$ . Thus $- \frac{\sqrt{3}}{2} = \frac{\frac{3}{\sqrt{2}}}{k}$ .

So $k = - 3 \sqrt{\frac{2}{3}} = - \sqrt{6}$ .

The tangent to the circle $x^{2} + y^{2} = 5$ at $(1, - 2)$ is

$x x_{1} + y y_{1} = r^{2}$ . So $x - 2 y = 5$ .

Now consider the second circle $x^{2} + y^{2} - 8 x + 6 y + 20 = 0$ . So the center is $(4, - 3)$ and radius is $\sqrt{5}$ .

Find the distance from the center to the line $x - 2 y - 5 = 0$ . So $\frac{| 4 + 6 - 5 |}{\sqrt{5}} = \sqrt{5}$ .

This equals the radius, so the line is tangent.

Put $x = 2 y + 5$ into the circle. So $(2 y + 5 - 4)^{2} + (y + 3)^{2} = 5$ . This gives $(2 y + 1)^{2} + (y + 3)^{2} = 5$ .

So $5 y^{2} + 10 y + 5 = 0 \Rightarrow y^{2} + 2 y + 1 = 0$ . So $(y + 1)^{2} = 0$ and $y = - 1$ . Then $x = 3$ .

Hence, the point of contact is $(3, - 1)$ .

The given family is $x^{2} + y^{2} - 2 x - 2 λ y - 8 = 0$ . Write it as $x^{2} + y^{2} - 2 x - 8 - 2 λ y = 0$ .

For fixed points, the equation must hold for all $λ$ . So $y = 0$ (for $x$ -axis) and $x^{2} - 2 x - 8 = 0$ .

Thus, $x = 4$ or $x = - 2$ . Hence the fixed points are $A (4, 0)$ and $B (- 2, 0)$ .

So $2 x + 2 y y^{'} - 2 - 2 λ y^{'} = 0$ . Thus $y^{'} = \frac{1 - x}{y - λ}$ .

At $A (4, 0)$ we get slope $\frac{3}{λ}$ . So tangent at $A$ is $y = \frac{3}{λ} (x - 4)$ .

At $B (- 2, 0)$ we get slope $- \frac{3}{λ}$ . So tangent at $B$ is $y = (- \frac{3}{λ}) (x + 2)$ .

Solving gives $x = 1$ and $y = - \frac{9}{λ}$ . This point lies on $x + 2 y + 5 = 0$ .

So $1 + 2 (- \frac{9}{λ}) + 5 = 0$ . Thus, $6 - \frac{18}{λ} = 0$ so $λ = 3$ .

Hence, the required circle is $x^{2} + y^{2} - 2 x - 6 y - 8 = 0$ .

Let $O$ be the center of the circle which is taken as the origin. Let $a$ be the radius of the circle. Now $A = (0, a), B = (0, - a)$ . Since $C D ∥ A B$ and $2 C D = A B$ .

Figure 11.13.

Let $C L ⊥ C D . C L = \frac{C D}{2} = \frac{a}{2}$

In $△ O L C, O L = \sqrt{O C^{2} - C L^{2}} = \frac{\sqrt{3} a}{2}$

$C = (\frac{\sqrt{3}}{2} a, \frac{a}{2}), D = (\frac{\sqrt{3}}{2} a, - \frac{a}{2})$ .

Now equation of the circle is $x^{2} + y^{2} = a^{2}$ and equation of tangent at $(0, - a)$ is $- a y = a^{2} \Rightarrow y = - a$

Equation of $A C$ is $y - a = \frac{0 - \frac{\sqrt{3}}{2} a}{a - \frac{a}{2}} (x - 0)$ . Solving this with $y = - a$ we get $E = (2 \sqrt{3} a, - a)$

Thus, $A E = 2. A B$ .

Let the center $C$ of the circle be taken as the origin and let $a$ be the radius of the given circle. Let $C X$ and $C Y$ be the $x$ and $y$ axes respectively. Let the two parallel tangents to the circle at $Q$ and $R$ be $y = a$ and $y = - a$ .

Equation of the circle is $x^{2} + y^{2} = a^{2} \dots$ (1) and equation of any other tangent at point $P$ be $y = m x + a \sqrt{1 + m^{2}} \dots$ (2)

Let $A$ and $B$ be the points of intersection of tangent (2) with the lines $y = a$ and $y = - a$ respectively, then

$A = (\frac{a - a \sqrt{1 + m^{2}}}{m}, a)$ and $B = (\frac{- a - a \sqrt{1 + m^{2}}}{m}, - a)$

Slope of $A C = \frac{a m}{a (1 - \sqrt{1 + m^{2}})} = m_{1}$ (let) and slope of $B C = \frac{a m}{a + \sqrt{1 + m^{2}}} = m_{2}$ (let)

$m_{1} m_{2} = - 1$ , hence, $∠ A C B = 90^{\circ}$ .

Figure 11.14.

Given $A$ is the origin, which is the center of circle I, $A X$ and $A Y$ are the $x$ and $y$ axes respectively. $B$ and $C$ are the center of the circles II and III respectively, and their radii are $3$ and $4$ respectively.

Figure 11.15.

Since circles I and II touch each other externally, therefore, $A B = 8$ and since circles I and III touch each other externally, therefore $A C = 9$ .

Let $B D, C E ⊥ x$ -axis. Then $A D = \sqrt{8^{2} - 3^{2}} = \sqrt{55}$ and $A E = \sqrt{9^{2} - 4^{2}} = \sqrt{65}$ .

Since both circles I and II touch $x$ -axis, therefore, $y = 0$ is their common tangent. Let $B C$ meet $y = 0$ at $H$ . Then one more tangent will pass through $H$ and $H$ will divided $B C$ internally or externally in the ratio $3 : 4$ according as circles II and III lie in different qudrants or in the same quadrant.

Case I: When circles II and III lie in the first and fourth quadrant respectively.

In this case $B = (\sqrt{55}, 3)$ and $C = (\sqrt{65}, - 4)$ .

From similar $△ B D H$ and $△ C E H, \frac{D H}{H E} = \frac{B D}{C E} = \frac{3}{4}$

Hence, $H$ divides $D E$ internally in the ratio $3 : 4$ . Thus, $H = (\frac{3 \sqrt{65} + 4 \sqrt{55}}{7}, 0)$

Equation of any line through $H$ will be $y = m (x - \frac{3 \sqrt{65} + 4 \sqrt{55}}{7})$

$\Rightarrow 7 m x - 7 y - m (3 \sqrt{65} + 4 \sqrt{55}) = 0 \dots$ (1)

If (1) is tangent to circle II then

$\frac{| 7 m \sqrt{55} - 7.3 - m (3 \sqrt{65} + 4 \sqrt{55}) |}{7 \sqrt{1 + m^{2}}} = 3$

$\Rightarrow m = 0, \frac{126 (\sqrt{55} - \sqrt{66})}{9 (71 - 10 \sqrt{143})}$

Thus, we have equation for common tangents.

Case II: When both circle II and III lie in the first qudrant.

In this case $B = (\sqrt{55}, 3)$ and $C = (\sqrt{65}, 4)$ .

One common tangent $y = 0$ meeting $B C$ at $H$ and $H$ will divide $B C$ extrenally in the ratio $3 : 4$ .

Thus, $H = (4 \sqrt{55} - 3 \sqrt{65}, 0)$ .

Now we can proceed like case 1 to find the other common tangent as well as case 3 when both the circles will lie in fourth quadrant.

Center of the circle is $(1, 2)$ and the point is $(2, 3)$ .

The equation of the normal will be equation of line passing through these points, which is

$y - 3 = \frac{2 - 3}{1 - 2} (x - 2) \Rightarrow x - y + 1 = 0$ .

The equation of the circle is $(x + 4) (x - 6) + (y - 4) (y + 1) = 0 \Rightarrow x^{2} + y^{2} - 2 x - 3 y - 28 = 0$

Putting $x = 0$ gives us $y^{2} - 3 y - 28 = 0 ∴ y = 7, - 4$

Hencem we take $A$ as $(0, 7)$ and $B$ as $(0, - 4)$ .

Equation of tangent at $A$ is $x .0 + y .7 - (x + 0) - \frac{3 (y + 7)}{2} - 28 = 0 \Rightarrow 2 x - 11 y = - 77$

Equation of tangent at $B$ is $x .0 + y (- 4) - (x + 0) - \frac{3 (y - 4)}{2} - 28 = 0 \Rightarrow 2 x + 11 y = - 44$

Solving the two tangents givens us $Q (- \frac{121}{4}, \frac{3}{2})$ .

Thus, area of the $△ A Q B = \frac{363}{8}$ .

Equation of the tangent is $x x_{1} + y y_{1} - 2 (x + x_{1}) - 3 (y + y_{1}) - 12 = 0$ .

Substituting $(x_{1}, y_{1}) = (- 1, - 1)$ gives $- x - y - 2 (x - 1) - 3 (y - 1) - 12 = 0$ .

Simplify to get $- 3 x - 4 y - 7 = 0$ . Hence, the equation of the tangent is $3 x + 4 y + 7 = 0$ .

The given circle is $x^{2} + y^{2} - 7 x - 5 y + 18 = 0$ . We find the tangent at $(4, 3)$ .

So $x x_{1} + y y_{1} - \frac{7 (x + x_{1})}{2} - \frac{5 (y + y_{1})}{2} + 18 = 0$ .

Substituting $(4, 3)$ gives $4 x + 3 y - \frac{7 (x + 4)}{2} - \frac{5 (y + 3)}{2} + 18 = 0$ .

Simplifying $x + y - 7 = 0$ .

Now find the tangent at $(3, 2)$ . So $3 x + 2 y - \frac{7 (x + 3)}{2} - \frac{5 (y + 2)}{2} + 18 = 0$ .

$\Rightarrow x + y - 5 = 0$ .

Both tangents have slope $- 1$ . Hence, they are parallel.

The given circle is $x^{2} + y^{2} = 169$ . The tangent at a point $(x_{1}, y_{1})$ on this circle is $x x_{1} + y y_{1} = 169$ .

At $(5, 12)$ the tangent is $5 x + 12 y = 169$ . At $(12, - 5)$ the tangent is $12 x - 5 y = 169$ .

From $5 x + 12 y = 169$ we get slope $- \frac{5}{12}$ . From $12 x - 5 y = 169$ we get slope $\frac{12}{5}$ .

Their product is $- 1$ so the tangents are perpendicular. Solving $5 x + 12 y = 169$ and $12 x - 5 y = 169$ gives $x = 17$ and $y = 7$ .

Hence, the point of intersection is $(17, 7)$ .

The equation of the tangent at $(α, β)$ is $x α + y β = r^{2}$ .

To find the intercepts, put $y = 0$ . Then $x α = r^{2}$ so $x = \frac{r^{2}}{α}$ .

So point $A$ is $(\frac{r^{2}}{α}, 0)$ .

Now put $x = 0$ . Then $y β = r^{2}$ so $y = \frac{r^{2}}{β}$ . So point $B$ is $(0, \frac{r^{2}}{β})$ .

Now the area of triangle $O A B$ is $\frac{1}{2} \times O A \times O B$ .

So area is $\frac{1}{2} \times \frac{r^{2}}{| α |} \times \frac{r^{2}}{| β |} = \frac{1}{2} \frac{r^{4}}{| α β |}$ .

The given circle is $x^{2} + y^{2} - 2 x - 4 y - 20 = 0 \Rightarrow (x - 1)^{2} + (y - 2)^{2} = 25$ .

So the center is $A (1, 2)$ . Tangent at $(x_{1}, y_{1})$ is $x x_{1} + y y_{1} - (x + x_{1}) - 2 (y + y_{1}) - 20 = 0$ .

Tangent at $(1, 7)$ is $x + 7 y - (x + 1) - 2 (y + 7) - 20 = 0 \Rightarrow y = 7$ .

Tangent at $(4, - 2)$ is $4 x - 2 y - (x + 4) - 2 (y - 2) - 20 = 0 \Rightarrow 3 x - 4 y - 20 = 0$ .

Solving the two tangents we get point of intersection as $C (16, 7)$ .

Split $A B C D$ into triangles $A B C$ and $A D C$ .

Triangle $A B C$ has base $B C = 15$ and height $5$ . So area is $\frac{1}{2} \times 15 \times 5 = \frac{75}{2}$ .

Triangle $A D C$ has base $D C = \sqrt{(16 - 4)^{2} + (7 + 2)^{2}} = 15$ and height $5$ . So area is $\frac{75}{2}$ .

Hence, total area is $75$ .

The given circle is $x^{2} + y^{2} - 2 x - 4 y + 3 = 0 \Rightarrow (x - 1)^{2} + (y - 2)^{2} = 2$ .

So the center is $(1, 2)$ and radius is $\sqrt{2}$ . The given line is $x + y - 5 = 0$ .

The distance from the center to the line is $\frac{| 1 + 2 - 5 |}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ .

This equals the radius, so the line touches the circle.

Put $y = 5 - x$ in the circle. So $(x - 1)^{2} + (3 - x)^{2} = 2$ .

Expanding gives $x^{2} - 2 x + 1 + x^{2} - 6 x + 9 = 2 \Rightarrow 2 x^{2} - 8 x + 10 = 2$ .

$2 x^{2} - 8 x + 8 = 0$ so $x^{2} - 4 x + 4 = 0 \Rightarrow (x - 2)^{2} = 0$ so $x = 2$ . Then $y = 3$ .

Hence, the point of contact is $(2, 3)$ .

The tangent to the circle $x^{2} + y^{2} = 5$ at $(1, - 2)$ is

Tangent at $(x_{1}, y_{1})$ is $x x_{1} + y y_{1} = r^{2}$ . So $x - 2 y = 5$ .

Now consider the second circle $x^{2} + y^{2} - 8 x + 6 y + 20 = (x - 4)^{2} + (y + 3)^{2} = 5$ .

So the center is $(4, - 3)$ and radius is $\sqrt{5}$ .

The distance from the center to the line $x - 2 y - 5 = 0$ . So $\frac{| 4 + 6 - 5 |}{\sqrt{5}} = \sqrt{5}$ .

This equals the radius, so the line is tangent.

Put $x = 2 y + 5$ into the circle. So $(2 y + 5 - 4)^{2} + (y + 3)^{2} = 5 \Rightarrow y^{2} + 2 y + 1 = 0$ .

So $(y + 1)^{2} = 0$ and $y = - 1$ . Then $x = 3$ .

Hence, the point of contact is $(3, - 1)$ .

The given circles are $x^{2} + y^{2} - 10 x + 4 y - 20 = 0$ and $x^{2} + y^{2} + 14 x - 6 y + 22 = 0$ .

First circle is $(x - 5)^{2} + (y + 2)^{2} = 49$ so center is $(5, - 2)$ and radius is $7$ .

Second circle is $(x + 7)^{2} + (y - 3)^{2} = 36$ so center is $(- 7, 3)$ and radius is $6$ .

Distance between centers is $\sqrt{(5 + 7)^{2} + (- 2 - 3)^{2}} = \sqrt{144 + 25} = 13$ .

Since $r_{1} + r_{2} = 7 + 6 = 13$ , the circles touch externally.

The point of contact lies on the line joining the centers and divides it in the ratio $7 : 6$ .

So using section formula $x = \frac{7 (- 7) + 6 (5)}{13} = - \frac{19}{13}$ and $y = \frac{7 (3) + 6 (- 2)}{13} = \frac{9}{13}$

Hence the point of contact is $(- \frac{19}{13}, \frac{9}{12})$ .

Tangent at $(x_{1}, y_{1})$ on first circle is $x x_{1} + y y_{1} - 5 (x + x_{1}) + 2 (y + y_{1}) - 20 = 0 \Rightarrow 19 x - 9 y + 110 = 0$ .

The given circle is $x^{2} + y^{2} = 2$ so the center is $(0, 0)$ and radius is $\sqrt{2}$ .

The given line is $y - x - 2 = 0$ .

The distance from the center to the line is $\frac{| 0 - 0 - 2 |}{\sqrt{2}} = \sqrt{2}$ .

This equals the radius, so the line touches the circle.

Put $y = x + 2$ in the circle. So $x^{2} + (x + 2)^{2} = 2$ . Thus, $(x + 1)^{2} = 0$ so $x = - 1$ .

Then $y = 1$ . Hence, the point of contact is $(- 1, 1)$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Its center is $(- g, - f)$ and radius is $\sqrt{g^{2} + f^{2} - c}$ .

For the line $l x + m y + n = 0$ to touch the circle, the distance from the center to the line must equal the radius.

So the condition is $\frac{| - l g - m f + n |}{\sqrt{l^{2} + m^{2}}} = \sqrt{g^{2} + f^{2} - c}$ .

Squaring both sides gives $(- l g - m f + n)^{2} = (l^{2} + m^{2}) (g^{2} + f^{2} - c)$ .

Point of contact is the foot of the perpendicular from the center to the line.

So the coordinates are $x = - g - l \frac{- l g - m f + n}{l^{2} + m^{2}}$ and $y = - f - \frac{m (- l g - m f + n)}{l^{2} + m^{2}}$ .

The given circle is $x^{2} + y^{2} = 10 x \Rightarrow (x - 5)^{2} + y^{2} = 25$ . So the center is $(5, 0)$ and radius is $5$ .

The line is $3 x + 4 y - k = 0$ . For tangency, the distance from the center to the line equals the radius.

So $\frac{| 15 - k |}{5} = 5 \Rightarrow | 15 - k | = 25$ . So $k = 40$ or $k = - 10$ .

Now find the point of contact using foot of perpendicular from $(5, 0)$ to the line.

For $k = 40, x = 5 - 3 \frac{15 - 40}{25} = 8$ and $y = 0 - 4 \frac{15 - 40}{25} = 4$ . So point is $(8, 4)$ .

For $k = - 10, x = 5 - 3 \frac{15 + 10}{25} = 2$ and $y = 0 - 4 \frac{15 + 10}{25} = - 4$ . So point is $(2, - 4)$ .

Hence, $k = 40$ or $k = - 10$ and the points of contact are $(8, 4)$ and $(2, - 4)$ .

The given circle is $x^{2} + y^{2} = 5$ so the center is $(0, 0)$ .

The normal at a point on a circle is the line joining the center to that point.

So the normal passes through $(0, 0)$ and $(1, 2)$ . The slope is $2$ .

Hence, the equation is $y = 2 x$ .

The given circle is $x^{2} + y^{2} = 2 x \Rightarrow (x - 1)^{2} + y^{2} = 1$ so the center is $(1, 0)$ .

The given line $x + 2 y = 3$ has slope $- \frac{1}{2}$ . So the required normal must also have slope $- \frac{1}{2}$ .

The normal to a circle passes through the center.

Hence, the normal is the line through $(1, 0)$ with slope $- \frac{1}{2}$ . So its equation is $y = - \frac{1}{2} (x - 1)$ .

Thus, $x + 2 y - 1 = 0$ .

Given circle is $x^{2} + y^{2} - 6 x - 10 y + k = 0$ . Let $P = (1, 4)$ . Since $P$ lies inside the circle $17 - 6 - 40 + k < 0 \Rightarrow k < 29$ .

Let $H$ be the center and $a$ the radius of the circle, then $H = (3, 5)$ and $a = \sqrt{34 - k}$ .

Since the circle neither cuts the $x$ -axis nor touches it $∴ a < | 5 | \Rightarrow k > 9$ .

Again since the circle neither cuts the $y$ -axis not touches it $∴ a < | 3 | \Rightarrow k > 25$

Combining the conditions we have $25 < k < 29$ .

Lenght of tangent is $\sqrt{5^{2} + 1^{2} + 6.5 - 4.1 - 3} = 7$ .

Given circles are $x^{2} + y^{2} - 2 λ x - c^{2} = 0 \dots$ (1) where $l a m b d a$ is a variable.

Let the three values of $λ$ be $λ_{1}, λ_{2}$ and $λ_{3}$ . Let $A, B$ and $C$ be the centers of the three circles respectively, then

$A = (λ_{1}, 0), B = (λ_{2}, 0)$ and $C = (λ_{3}, 0)$ . If $O (0, 0)$ be the origin, then

$O A = | λ_{1} |, O B = | λ_{2} |$ and $O C = | λ_{3} |$ .

Given that $| λ_{1} |, | λ_{2} |, | λ_{3} |$ are in G.P. $∴ | λ_{2} |^{2} = | λ_{1} | | λ_{3} |$

Equation of another circle is $x^{2} + y^{2} = c^{2}$ . Let $P (α, β)$ be any circle on this point, then

$α^{2} + β^{2} - c^{2} = 0$

Lengths of tangents from $P$ to the three circles are $p_{1} = \sqrt{α^{2} + β^{2} - 2 λ_{1} α - c^{2}} = \sqrt{- 2 λ_{1} α}, p_{2} = \sqrt{- 2 λ_{2} α}$ , and $p_{3} = \sqrt{- 2 λ_{3} α}$

$p_{1} p_{3} = \sqrt{4 λ_{1} λ_{3} α^{2}}$ . We see that $λ_{1} λ_{3} > 0 \Rightarrow | λ_{1} | | λ_{3} | = λ_{1} λ_{3}$

Thus, $p_{2}^{2} = p_{1} p_{3}$ , and hence, $p_{1}, p_{2}, p_{3}$ are in G.P.

Let $P = (α, β)$ . Given that the lengths of the tangents are equal, therefore,

$\sqrt{α^{2} + β^{2} + α - 3} = \sqrt{α^{2} + β^{2} - \frac{5}{3} α + β} = \sqrt{α^{2} + β^{2} + 2 α + \frac{7}{4} β + \frac{9}{4}}$

Solving we get $α = 0, β = - 3 ∴ P = (0, - 3)$ .

Let the equation of the required circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . It passes through $(0, - 3)$ , therefore,

$- 6 f + c + 9 = 0$

Equation of the tangent to the circle at $(6, - 1)$ is $6 x - y + g (x + 6) + f (y - 1) + c = 0$

Given that the equation of the tangent is $x + y - 5 = 0$ .

Comparing coefficients we have $g = - \frac{7}{2}, f = \frac{7}{2} \Rightarrow c = 12$

Thus, equation of the circle is $x^{2} + y^{2} - 7 x + 7 y + 12 = 0$ .

The length of tangent from $(f, g)$ to $x^{2} + y^{2} = 6$ is $\sqrt{f^{2} + g^{2} - 6}$ .

The length of tangent from $(f, g)$ to $x^{2} + y^{2} + 3 x + 3 y = 0$ is $\sqrt{f^{2} + g^{2} + 3 f + 3 g}$ .

Given the first is twice the second. So $\sqrt{f^{2} + g^{2} - 6} = 2 \sqrt{f^{2} + g^{2} + 3 f + 3 g}$ .

So $f^{2} + g^{2} - 6 = 4 (f^{2} + g^{2} + 3 f + 3 g)$ .

Simplify to get $0 = 3 f^{2} + 3 g^{2} + 12 f + 12 g + 6$ . So $f^{2} + g^{2} + 4 f + 4 g + 2 = 0$ .

The length of the tangent from $(f, g)$ to the circle $x^{2} + y^{2} = 4$ is $\sqrt{f^{2} + g^{2} - 4}$ .

The second circle is $x^{2} + y^{2} = 4 x$ which is $(x - 2)^{2} + y^{2} = 4$ .

So the length of the tangent from $(f, g)$ to this circle is $\sqrt{(f - 2)^{2} + g^{2} - 4}$ .

Given $\sqrt{f^{2} + g^{2} - 4} = 4 \sqrt{(f - 2)^{2} + g^{2} - 4}$ .

So $f^{2} + g^{2} - 4 = 16 f^{2} + 16 g^{2} - 64 f \Rightarrow 15 f^{2} + 15 g^{2} - 64 f + 4 = 0$ .

Let $(x_{1}, y_{1})$ be any point on the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

So $x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c = 0$ .

The length of the tangent from $(x_{1}, y_{1})$ to the circle $x^{2} + y^{2} + 2 g x + 2 f y + c_{1} = 0$ is

$\sqrt{x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c_{1}}$ .

Using the first relation, substitute $x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} = - c$ .

So the length becomes $\sqrt{- c + c_{1}}$ .

Let the required point be $(x, y)$ . The length of the tangent from $(x, y)$ to $x^{2} + y^{2} = 1$ is

$\sqrt{x^{2} + y^{2} - 1}$ . For the circle $x^{2} + y^{2} - 8 x + 15 = 0$ it is $\sqrt{x^{2} + y^{2} - 8 x + 15}$ .

For the circle $x^{2} + y^{2} + 10 y + 24 = 0$ it is $\sqrt{x^{2} + y^{2} + 10 y + 24}$ .

Given $x^{2} + y^{2} - 1 = x^{2} + y^{2} - 8 x + 15 \Rightarrow x = 2$ .

Again $x^{2} + y^{2} - 1 = x^{2} + y^{2} + 10 y + 24 \Rightarrow y = - \frac{5}{2}$ .

Hence, the required point is $(2, - \frac{5}{2})$ .

Let the required point be $(x, y)$ . The length of the tangent from $(x, y)$ to $x^{2} + y^{2} - 4 x + 7 = 0$ is

$\sqrt{x^{2} + y^{2} - 4 x + 7}$ . For the circle $2 x^{2} + 2 y^{2} - 3 x + 5 y + 9 = 0$ divide by $2$ .

So it becomes $x^{2} + y^{2} - \frac{3}{2} x + \frac{5}{2} y + \frac{9}{2} = 0$ .

Hence, the length is $\sqrt{x^{2} + y^{2} - \frac{3}{2} x + \frac{5}{2} y + \frac{9}{2}}$ .

For the circle $x^{2} + y^{2} + y = 0$ the length is $\sqrt{x^{2} + y^{2} + y}$ .

Equating first and second. So $x^{2} + y^{2} - 4 x + 7 = x^{2} + y^{2} - \frac{3}{2} x + \frac{5}{2} y + \frac{9}{2}$ .

Thus, $- 5 x - 5 y + 5 = 0$ so $x + y = 1$ .

Now equate first and third. So $x^{2} + y^{2} - 4 x + 7 = x^{2} + y^{2} + y$ . So $y = - 4 x + 7$ .

Solve with $x + y = 1$ . So $x - 4 x + 7 = 1$ .

$\Rightarrow x = 2$ and $y = - 1$ . Hence, the point is $(2, - 1)$ .

So length is $\sqrt{4 + 1 - 8 + 7} = 2$ .

Let the point be $(x, y)$ . For the circle $x^{2} + y^{2} + 2 g_{i} x + 5 = 0$ , the length of the tangent from $(x, y)$ is $t_{i}^{2} = x^{2} + y^{2} + 2 g_{i} x + 5$ .

Now consider $(g_{2} - g_{3}) t_{1}^{2} + (g_{3} - g_{1}) t_{2}^{2} + (g_{1} - g_{2}) t_{3}^{2}$ .

Substitute $t_{i}^{2}$ . So it becomes $(g_{2} - g_{3}) (x^{2} + y^{2} + 2 g_{1} x + 5) +$
$(g_{3} - g_{1}) (x^{2} + y^{2} + 2 g_{2} x + 5) + (g_{1} - g_{2}) (x^{2} + y^{2} + 2 g_{3} x + 5)$ .

The coefficient of $(x^{2} + y^{2} + 5)$ is $(g_{2} - g_{3}) + (g_{3} - g_{1}) + (g_{1} - g_{2}) = 0$ .

Now consider the remaining terms.

So we get $2 x [(g_{2} - g_{3}) g_{1} + (g_{3} - g_{1}) g_{2} + (g_{1} - g_{2}) g_{3}]$ .

Expanding inside. So $g_{1} g_{2} - g_{1} g_{3} + g_{2} g_{3} - g_{1} g_{2} + g_{1} g_{3} - g_{2} g_{3} = 0$ .

Hence, the whole expression is $2$ .

Let the point be $(x, y)$ . The length of the tangent from $(x, y)$ to $x^{2} + y^{2} = a^{2}$ is $\sqrt{x^{2} + y^{2} - a^{2}}$ .

The second circle is $(x - a)^{2} + y^{2} = a^{2}$ . So the length of the tangent from $(x, y)$ to this circle is $\sqrt{(x - a)^{2} + y^{2} - a^{2}}$ .

Given the first is four times the second. So $\sqrt{x^{2} + y^{2} - a^{2}} = 4 \sqrt{(x - a)^{2} + y^{2} - a^{2}}$ .

$\Rightarrow x^{2} + y^{2} - a^{2} = 16 ((x - a)^{2} + y^{2} - a^{2}) \Rightarrow x^{2} + y^{2} - a^{2} = 16 (x^{2} + y^{2} - 2 a x)$ .

So $0 = 15 x^{2} + 15 y^{2} - 32 a x + a^{2}$ . Hence, the point lies on the required circle.

The equation of the pair of tangents from $(0, 1)$ is given by $T^{2} = S S_{1}$ .

Here $S = x^{2} + y^{2} - 2 x + 4 y$ and $S_{1} = 0^{2} + 1^{2} - 2 (0) + 4 (1) = 5$ .

Now $T$ is $x \times 0 + y \times 1 - (x + 0) + 2 (y + 1)$ . So $T = y - x + 2 y + 2 = - x + 3 y + 2$ .

Thus the equation is $(- x + 3 y + 2)^{2} = 5 (x^{2} + y^{2} - 2 x + 4 y)$

$\Rightarrow 3 x^{2} - 2 y^{2} + 3 x y - 3 x + 4 y - 2 = 0$ .

Let $(x_{1}, y_{1})$ be any point on the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . So $x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c = 0$ .

The second circle is $x^{2} + y^{2} + 2 g x + 2 f y + c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α = 0$ .

The length of the tangent from $(x_{1}, y_{1})$ to the second circle is
$\sqrt{x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α}$ .

Using the first relation, substitute $x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} = - c$ .

So the length becomes $\sqrt{- c + c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α}$ .

This simplifies to $\sqrt{(g^{2} + f^{2} - c) \cos^{2} α}$ . So the tangent length is $\sqrt{g^{2} + f^{2} - c} \cos α$ .

The radius of the second circle is $\sqrt{g^{2} + f^{2} - (c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α)}$ .

$= \sqrt{(g^{2} + f^{2} - c) \sin^{2} α}$ .

Let $θ$ be the angle between the tangents. Then $\tan \frac{θ 2}{=} \frac{r}{d}$ where $d$ is the tangent length.

So $\tan \frac{θ}{2} = \tan α$ . Hence, $θ = 2 α$ .

The given circle is $x^{2} + y^{2} = 25$ so the center is $(0, 0)$ and radius is $5$ .

Let the tangent from $(1, - 7)$ have slope $m$ . So its equation is $y + 7 = m (x - 1)$ .

This gives $m x - y - m - 7 = 0$ .

For tangency, the distance from the center to the line equals the radius. So $\frac{| - m - 7 |}{\sqrt{m^{2} + 1}} = 5$ .

$\Rightarrow m = \frac{4}{3}$ or $m = - \frac{3}{4}$ .

Hence, the tangents are $y + 7 = \frac{4}{3} (x - 1)$ and $y + 7 = - \frac{3}{4} (x - 1)$ .

Their product of slopes is $- 1$ so they are perpendicular.

The given circle is $x^{2} + y^{2} = 16$ . From the point $(9, 0)$ the length of the tangent is

$\sqrt{9^{2} - 16} = \sqrt{65}$ which is real. Hence, two tangents can be drawn.

The equation of the pair of tangents is given by $T^{2} = S S_{1}$ .

Here $S = x^{2} + y^{2} - 16$ and $S_{1} = 81 - 16 = 65$ .

Now $T = x x_{1} + y y_{1} - 16 = 9 x - 16$ .

So the equation is $(9 x - 16)^{2} = 65 (x^{2} + y^{2} - 16)$ .

Thus, $16 x^{2} - 65 y^{2} - 288 x + 1296 = 0$ .

So $\tan θ = 2 \frac{\sqrt{0 - (16) (- 65)}}{16 - 65} = - \frac{8 \sqrt{65}}{49}$ .

The given circle is $x^{2} + y^{2} = 25$ so the center is $(0, 0)$ and radius is $5$ .

Let the tangent through $(7, 1)$ have slope $m$ .

So its equation is $y - 1 = m (x - 7)$ . This gives $m x - y - 7 m + 1 = 0$ .

For tangency, the distance from the center to the line equals the radius.

So $\frac{| - 7 m + 1 |}{\sqrt{m^{2} + 1}} = 5$ . Solve to get $m = \frac{4}{3}$ or $m = - \frac{3}{4}$ .

Hence, the tangents are $y - 1 = \frac{4}{3} (x - 7)$ and $y - 1 = - \frac{3}{4} (x - 7)$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + k^{2} = 0$ .

The equation of the pair of tangents from the origin is given by $T^{2} = S S_{1}$ .

Here $S = x^{2} + y^{2} + 2 g x + 2 f y + k^{2}$ and $S_{1} = k^{2}$ . Now $T = g x + f y + k^{2}$ .

So the equation is ${(g x + f y + k^{2})}^{2} = k^{2} (x^{2} + y^{2} + 2 g x + 2 f y + k^{2})$ .

This is the required pair of tangents.

Now find the intercept on the line $y = h$ . Substitute $y = h$ .

So ${(g x + f h + k^{2})}^{2} = k^{2} (x^{2} + h^{2} + 2 g x + 2 f h + k^{2})$ .

The intercept is the distance between the two roots.

So length is $\frac{2 \sqrt{(g^{2} - k^{2}) (h^{2} + k^{2} + 2 f h)}}{| k^{2} - g^{2} |}$ .

Using the relation $g^{2} + f^{2} - k^{2} = r^{2}$ simplify the expression.

This reduces to $\frac{2 h k r}{k^{2} - g^{2}}$ .

Hence, the intercept is $\frac{2 h k}{k^{2} - g^{2}}$ times the radius.

The given circle is $x^{2} + y^{2} + 6 x + 8 y - 11 = 0$ . Let the midpoint of the chord be $(1, - 1)$ .

The chord whose midpoint is $(x_{1}, y_{1})$ is given by $T = S_{1}$ .

Here $S = x^{2} + y^{2} + 6 x + 8 y - 11$ and $S_{1} = 1^{2} + (- 1)^{2} + 6 (1) + 8 (- 1) - 11 = - 11$ .

Now $T = x x_{1} + y y_{1} + 3 (x + x_{1}) + 4 (y + y_{1}) - 11$ .

Substitute $(1, - 1)$ . So $T = x - y + 3 (x + 1) + 4 (y - 1) - 11$ .

Thus, the chord is $T = S_{1}$ so $4 x + 3 y - 12 = - 11$ .

Hence, the equation is $4 x + 3 y - 1 = 0$ .

The given circle is $x^{2} + y^{2} + 6 x + 8 y + 9 = 0$ . Let the midpoint be $(- 2, - 3)$ .

The chord whose midpoint is $(x_{1}, y_{1})$ is given by $T = S_{1}$ .

Here $S = x^{2} + y^{2} + 6 x + 8 y + 9$ and $S_{1} = (- 2)^{2} + (- 3)^{2} + 6 (- 2) + 8 (- 3) + 9 = - 14$ .

Now $T = x x_{1} + y y_{1} + 3 (x + x_{1}) + 4 (y + y_{1}) + 9$ .

Substitute $(x_{1}, y_{1}) = (- 2, - 3)$ . So $T = - 2 x - 3 y + 3 (x - 2) + 4 (y - 3) + 9$ .

Simplify to get $x + y - 9$ . Thus, the chord is $T = S_{1}$ so $x + y - 9 = - 14$ .

Hence, the equation is $x + y + 5 = 0$ .

The given circle is $x^{2} + y^{2} + 4 x - 2 y - 3 = 0$ . So the center is $(- 2, 1)$ .

The given line is $y = x + 2$ or $x - y + 2 = 0$ .

The midpoint of the chord is the foot of the perpendicular from the center to the line.

Using the formula for foot of perpendicular from $(x_{1}, y_{1})$ to $a x + b y + c = 0$ ,

$x = x_{1} - \frac{a (a x_{1} + b y_{1} + c)}{a^{2} + b^{2}}$ and $y = y_{1} - \frac{b (a x_{1} + b y_{1} + c)}{a^{2} + b^{2}}$ .

Here $(x_{1}, y_{1}) = (- 2, 1)$ and $a = 1, b = - 1, c = 2$ .

$a x_{1} + b y_{1} + c = - 2 - 1 + 2 = - 1$ .

So $x = - 2 - 1 (- 1) / 2 = - \frac{3}{2}$ and $y = 1 - (- 1) (- 1) / 2 = \frac{1}{2}$ .

Hence, the midpoint is $(- \frac{3}{2}, \frac{1}{2})$ .

The given circle is $x^{2} + y^{2} - 2 x + 4 y + 7 = 0$ . The chord of contact from a point $(x_{1}, y_{1})$ is given by $T = 0$ .

Here $T = x x_{1} + y y_{1} - (x + x_{1}) + 2 (y + y_{1}) + 7$ .

Substitute $(x_{1}, y_{1}) = (1, 2)$ . So $T = x + 2 y - (x + 1) + 2 (y + 2) + 7$ .

$\Rightarrow 4 y + 10 = 0 \Rightarrow 2 y + 5 = 0$ .

Given circle is $x^{2} + y^{2} = a^{2} \dots$ (1)

Figure 11.16.

The equation of the chord of contact $A B$ of tangents drawn from $P (h, k)$ to the circle (1) is $x h + y k = a^{2}$ .

We have to find the area of $△ P A B$ . From $P (h, k)$ draw $P L ⊥ A B$ . Now

$P L = \frac{h^{2} + k^{2} - a^{2}}{\sqrt{h^{2} + k^{2}}}$

Also, $P A = \sqrt{h^{2} + k^{2} - a^{2}}$

$A L^{2} = A P^{2} - P L^{2} = \frac{a \sqrt{h^{2} + k^{2} - a^{2}}}{\sqrt{h^{2} + k^{2}}}$

$Δ_{P A B} = \frac{1}{2} . A B . P L = A L . P L = \frac{a {(h^{2} + k^{2} - a^{2})}^{3 / 2}}{h^{2} + k^{2}}$ .

Given circles are $x^{2} + y^{2} = a^{2} \dots$ (1), $x^{2} + y^{2} = b^{2} \dots$ (2), and $x^{2} + y^{2} = c^{2} \dots$ (3).

Let $P (α, β)$ be any point on (1), then $α^{2} + β^{2} = a^{2} \dots$ (4)

Equation of the chord of contact of the tangents from $P (α, β)$ to (2) is

$x α + y β - b^{2} = 0$ .

This chord of contact is tangent to (3), therefore,

$\frac{| 0. α + 0. β - b^{2} |}{\sqrt{α^{2} + β^{2}}} = c \Rightarrow b^{2} = a c$ , and hence, $a, b, c$ are in G.P.

Common chord of the circles is $5 x - 3 y - 10 = 0$ . Let this meet the circles at $A$ and $B$ . Let the tangents to first circle at $A$ and $B$ meet at $P (α, β)$ , then $A B$ will be the chord of contact of the tangents to the circle from $P$ , therefore, equation of $A B$ will be

$x α + y β - 12 = 0$

The two obtained equations are same. Comparing coefficients we have $\frac{α}{5} = \frac{β}{- 3} = \frac{12}{10}$ , which yields

$α = 6, β = - \frac{18}{5}$ .

The given circle is $x^{2} + y^{2} + 2 x - 3 = 0$ . The chord of contact from a point $(x_{1}, y_{1})$ is given by $T = 0$ .

Here $T = x x_{1} + y y_{1} + (x + x_{1}) - 3$ . Substitute $(x_{1}, y_{1}) = (- 3, 2)$ .

So $T = - 3 x + 2 y + (x - 3) - 3$ .

Hence, the chord of contact is $x - y + 3 = 0$ .

The given circle is $x^{2} + y^{2} = 25$ . The chord of contact from a point $(x_{1}, y_{1})$ is given by $x x_{1} + y y_{1} = 25$ .

Substitute $(x_{1}, y_{1}) = (5, 3)$ . So the equation becomes $5 x + 3 y = 25$ .

The given circle is $x^{2} + y^{2} - 4 x + 3 y - 1 = 0$ . The line is $2 x + y + 12 = 0$ .

Let the points of intersection be $P$ and $Q$ . The intersection of tangents at $P$ and $Q$ is given by the pole of the line.

So we find the pole of $2 x + y + 12 = 0$ with respect to the circle.

For the circle, the pole of $l x + m y + n = 0$ is $\frac{(- 2 g l - m n, - 2 f m - l n)}{l^{2} + m^{2}}$ .

Here $g = - 2$ and $f = \frac{3}{2}$ . $\Rightarrow x = \frac{- 2 (- 2) (2) - (1) (12)}{5} = - \frac{4}{5}$ and $y = \frac{- 2 (3 / 2) (1) - (2) (12)}{5} = - \frac{27}{5}$ .

Let the given circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Let the two points be $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ .

Since they are conjugate with respect to the circle, we have $x_{1} x_{2} + y_{1} y_{2} + g (x_{1} + x_{2}) + f (y_{1} + y_{2}) + c = 0$ .

The lengths of tangents are $t_{1}^{2} = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c$ and $t_{2}^{2} = x_{2}^{2} + y_{2}^{2} + 2 g x_{2} + 2 f y_{2} + c$ .

$t_{1}^{2} + t_{2}^{2} = x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} + 2 g (x_{1} + x_{2}) + 2 f (y_{1} + y_{2}) + 2 c$ .

Now consider the square of the distance between the points.

So $P Q^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}$ $= x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} - 2 (x_{1} x_{2} + y_{1} y_{2})$ .

Using the conjugate condition, $x_{1} x_{2} + y_{1} y_{2} = - g (x_{1} + x_{2}) - f (y_{1} + y_{2}) - c$ .

$∴ P Q^{2} = x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} + 2 g (x_{1} + x_{2}) + 2 f (y_{1} + y_{2}) + 2 c$ .

Thus, $P Q^{2} = t_{1}^{2} + t_{2}^{2}$ . Hence, $P Q = \sqrt{t_{1}^{2} + t_{2}^{2}}$ .

The given circle is $x^{2} + y^{2} = 25$ . Let the point be $P (4, 6)$ .

The chord of contact from $P$ is $4 x + 6 y = 25$ .

The area of the triangle formed by the two tangents and their chord of contact is

$Δ = \frac{r^{2} \times \sqrt{S_{1}}}{distance from center to chord}$ .

Here $S_{1} = 4^{2} + 6^{2} - 25 = 27$ . So $\sqrt{S_{1}} = 3 \sqrt{3}$ .

The distance from the center $(0, 0)$ to the chord $4 x + 6 y - 25 = 0$ is

$= \frac{25 \times 3 \sqrt{3} \times 2 \sqrt{13}}{25} = 6 \sqrt{39}$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ be the points of contact of tangents from the origin.

The tangent at $(x_{1}, y_{1})$ is $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$ .

Since this tangent passes through the origin, substitute $(0, 0)$ . So $g x_{1} + f y_{1} + c = 0$ .

Also since $(x_{1}, y_{1})$ lies on the circle, $x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c = 0$ .

Using $g x_{1} + f y_{1} = - c$ , we get $x_{1}^{2} + y_{1}^{2} - c = 0 \Rightarrow x_{1}^{2} + y_{1}^{2} = c$ . Similarly $x_{2}^{2} + y_{2}^{2} = c$ .

Now consider the circle $x^{2} + y^{2} + g x + f y = 0$ . Substitute $(0, 0)$ and it satisfies the equation.

Now substitute $(x_{1}, y_{1})$ . Using $x_{1}^{2} + y_{1}^{2} = c$ and $g x_{1} + f y_{1} = - c$ , we get $c - c = 0$ .

So $(x_{1}, y_{1})$ lies on it. Similarly $(x_{2}, y_{2})$ lies on it. Hence this circle passes through $O, P, Q$ .

Therefore, it is the circumcircle of $△ O P Q$ .

The given circle is $x^{2} + y^{2} = 25$ . Let the tangent through $(7, 1)$ have slope $m$ .

So its equation is $y - 1 = m (x - 7)$ . This gives $m x - y - 7 m + 1 = 0$ .

For tangency, the distance from the center $(0, 0)$ to the line equals the radius $5$ .

So $\frac{| - 7 m + 1 |}{\sqrt{m^{2} + 1}} = 5$ . Solve to get $m = \frac{4}{3}$ or $m = - \frac{3}{4}$ .

Hence the tangents are $y - 1 = \frac{4}{3} (x - 7)$ and $y - 1 = - \frac{3}{4} (x - 7)$ .

For $m = \frac{4}{3}$ , the tangent is $4 x - 3 y - 25 = 0$ .

Solve with the circle. So $x^{2} + \frac{(4 x - 25)^{2}}{9} = 25$ . This gives $x = 4$ and $y = 3$ .

For $m = - \frac{3}{4}$ , the tangent is $3 x + 4 y - 25 = 0$ . Solve with the circle. So $x^{2} + \frac{(25 - 3 x)^{2}}{16} = 25$ .

This gives $x = 3$ and $y = - 4$ . Hence the points of contact are $(4, 3)$ and $(3, - 4)$ .

Equation of the polar is $x .2 + y . (- 1) - 3 \frac{x + 2}{2} + 4 \frac{y - 1}{2} - 8 = 0 \Rightarrow x + 2 y - 26 = 0$ .

Let $P (α, β)$ be the pole of the given line w.r.t. the given circle. Equation of polar is

$(α + 2) x + (β + 3) y + 2 α + 3 β + 9 = 0$

Comparing with the given line $\frac{α + 2}{3} = \frac{β + 3}{5} = \frac{2 α + 3 β + 9}{17}$

$\Rightarrow α = 1, β = 2$ . So the required pole is $(1, 2)$ .

Given circles are $x^{2} + y^{2} + 6 y + 5 = 0 \dots$ (1) and $x^{2} + y^{2} + 2 x + 8 y + 5 = 0 \dots$ (2). Let $P = (1, - 2)$ .

Polar of the point $(1, - 2)$ w.r.t circle (1) is given by $x + y . (- 2) + 3 (y - 2) + 5 = 0 \Rightarrow x + y - 1 = 0 \dots$ (3)

Polar of this point w.r.t circle (2) is given by $x + y . (- 2) + x + 1 + 4 (y - 2) + 5 = 0 \Rightarrow x + y - 1 = 0$ .

Thus, polars are same. Let $Q (α, β)$ be another point for which the polars are same. The polars of this point w.r.t. given circles are

$x α + y β + 3 (y + β) + 5 = 0$ and $x α + y β + (x + α) + 4 (y + β) + 5 = 0$

These two lines are same. Thus, comparing coefficients gives us

$\frac{α + 1}{α} = \frac{β + 4}{β + 3} = \frac{α + 4 β + 5}{3 β + 5}$

Solving this gives us two points one of whihc is the given point and another point is $(2, - 1)$ .

Let the circle be $x^{2} + y^{2} = a^{2}$ and points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ .

Polars of $A$ and $B$ will be $x x_{1} + y y_{1} - a^{2} = 0$ and $x x_{2} + y y_{2} - a^{2} = 0$ .

$\frac{A M}{(B N} = \frac{| x_{1} x_{2} + y_{1} y_{2} - a^{2} |}{\sqrt{x_{2}^{2} + y_{2}^{2}}} = \frac{| x_{2} x_{2} + y_{2} y_{2} - a^{2} |}{\sqrt{x_{1}^{2} + y_{2}^{2}}} = \frac{\sqrt{x_{1}^{2} + y_{1}^{2}}}{\sqrt{x_{2}^{2} + y_{2}^{2}}} = \frac{C A}{C B}$ .

First find the point of intersection of the lines $4 x - y = 11$ and $x - 2 y = 1$ .

From the second equation we get $x = 1 + 2 y$ . Substitute in the first equation.

So $4 (1 + 2 y) - y = 11$ . Thus, $x = 3$ . So the point is $(3, 1)$ .

Now find the polar with respect to $x^{2} + y^{2} = 7$ .

The polar of $(x_{1}, y_{1})$ is $x x_{1} + y y_{1} = 7$ .

So the required equation is $3 x + y = 7$ .

The given circle is $2 x^{2} + 2 y^{2} = 11 \Rightarrow x^{2} + y^{2} = \frac{11}{2}$ .

So the polar is $4 x - y = \frac{11}{2}$ .

The given circle is $x^{2} + y^{2} - 8 x + 6 y + 4 = 0$ . The polar of a point $(x_{1}, y_{1})$ is given by $T = 0$ .

So $T = x x_{1} + y y_{1} - 4 (x + x_{1}) + 3 (y + y_{1}) + 4$ .

Substitute $(x_{1}, y_{1}) = (1, - 5)$ . So $T = x - 5 y - 4 (x + 1) + 3 (y - 5) + 4$ .

Hence, the polar is $3 x + 2 y + 15 = 0$ .

The polar of $(p, q)$ with respect to $x^{2} + y^{2} = a^{2}$ is $p x + q y = a^{2}$ .

For this line to touch the circle $(x - c)^{2} + (y - d)^{2} = b^{2}$ , the distance from its center $(c, d)$ to the line must equal the radius $b$ .

So $\frac{| p c + q d - a^{2} |}{\sqrt{p^{2} + q^{2}}} = b \Rightarrow {(p c + q d - a^{2})}^{2} = b^{2} (p^{2} + q^{2})$

$\Rightarrow b^{2} (p^{2} + q^{2}) = {(a^{2} - c p - d q)}^{2}$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

The polar of the origin is obtained by putting $(0, 0)$ in $T$ .

So the polar is $g x + f y + c = 0$ .

For this line to touch the circle $x^{2} + y^{2} = a^{2}$ , the distance from the center $(0, 0)$ to the line must equal the radius $a$ .

So $\frac{| c |}{\sqrt{g^{2} + f^{2}}} = a$ . Hence, $c^{2} = a^{2} (g^{2} + f^{2})$ .

Let the given line be $l x + m y + n = 0$ . The pole of this line with respect to $x^{2} + y^{2} = c^{2}$ is $(- \frac{c^{2} l}{n}, - \frac{c^{2} m}{n})$ .

Given this point lies on $x^{2} + y^{2} = 9 c^{2}$ . So $\frac{c^{4} l^{2}}{n^{2}} + \frac{c^{4} m^{2}}{n^{2}} = 9 c^{2}$ .

Thus, $c^{4} (l^{2} + m^{2}) = 9 c^{2} n^{2}$ . So $n^{2} = (\frac{c^{2}}{0} (l^{2} + m^{2})$ .

Now consider the circle $9 x^{2} + 9 y^{2} = c^{2}$ . Its center is $(0, 0)$ and radius is $\frac{c}{3}$ .

The distance from the center to the line is $\frac{| n |}{\sqrt{l^{2} + m^{2}}}$ .

Using the relation above, this becomes $\frac{c}{3}$ .

So the distance equals the radius. Hence the line is tangent to the circle $9 x^{2} + 9 y^{2} = c^{2}$ .

The given circle is $2 x^{2} + 2 y^{2} - 3 x + 5 y - 7 = 0 \Rightarrow x^{2} + y^{2} - \frac{3}{2} x + \frac{5}{2} y - \frac{7}{2} = 0$ . Thus, $g = - \frac{3}{4}$ and $f = \frac{5}{4}$ .

The pole $(x_{1}, y_{1})$ of the line $9 x + y - 28 = 0$ satisfies that this line is the polar of $(x_{1}, y_{1})$ .

So write $T = 0$ . Thus, $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$ .

So coefficient of $x$ is $x_{1} + g$ and coefficient of $y$ is $y_{1} + f$ .

Hence, the equation becomes $(x_{1} + g) x + (y_{1} + f) y + (g x_{1} + f y_{1} + c) = 0$ .

Compare with $9 x + y - 28 = 0$ . So $x_{1} + g = 9$ and $y_{1} + f = 1$ .

Substitute values. So $x_{1} - \frac{3}{4} = 9$ hence $x_{1} = \frac{39}{4}$ . And $y_{1} + \frac{5}{4} = 1$ hence $y_{1} = - \frac{1}{4}$ .

So the pole is $(\frac{39}{4}, - \frac{1}{4})$ . So it becomes $\frac{9}{4} x + \frac{1}{4} y - 7 = 0$ .

Now compare again. So $x_{1} + g = \frac{9}{4}$ gives $x_{1} = 3$ . And $y_{1} + f = \frac{1}{4}$ gives $y_{1} = - 1$ .

The given circle is $x^{2} + y^{2} - 7 x + 5 y - 1 = 0$ . Compare with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

So $g = - \frac{7}{2}, f = \frac{5}{2}, c = - 1$ . Let the pole be $(x_{1}, y_{1})$ . The polar of $(x_{1}, y_{1})$ is

$(x_{1} + g) x + (y_{1} + f) y + (g x_{1} + f y_{1} + c) = 0$ . This must represent the line $2 x - y + 10 = 0$ .

So equate coefficients with a factor $k : x_{1} + g = 2 k, y_{1} + f = - k$ and $g x_{1} + f y_{1} + c = 10 k$ .

Substitute $g = - \frac{7}{2}, f = \frac{5}{2}, c = - 1$ .

So $x_{1} - \frac{7}{2} = 2 k$ hence $x_{1} = 2 k + \frac{7}{2}$ and $y_{1} + \frac{5}{2} = - k$ hence $y_{1} = - k - \frac{5}{2}$ .

Substitute into the third equation: $- \frac{7}{2} (2 k + \frac{7}{2}) + \frac{5}{2} (- k - \frac{5}{2}) - 1 = 10 k$ .

Expand: $- 7 k - \frac{49}{4} - \frac{5 k}{2} - \frac{25}{4} - 1 = 10 k \Rightarrow - 7 k - \frac{5 k}{2} - \frac{37}{2} - 1 = 10 k$ .

Write $- 1 = - \frac{2}{2} : - 7 k - \frac{5 k}{2} - \frac{39}{2} = 10 k \Rightarrow k = - 1$ .

Now $x_{1} = 2 k + \frac{7}{2} = - 2 + \frac{7}{2} = \frac{3}{2}$ and $y_{1} = - k - \frac{5}{2} = 1 - \frac{5}{2} = - \frac{3}{2}$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Let the pole of the line $a x + b y + c = 0$ be $(x_{1}, y_{1})$ . The polar of $(x_{1}, y_{1})$ is

$x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0 \Rightarrow (x_{1} + g) x + (y_{1} + f) y + (g x_{1} + f y_{1} + c) = 0$ .

This must represent the same line as $a x + b y + c = 0$ .

So coefficients are proportional. Let the factor be $k$ .

$x_{1} + g = a k$ and $y_{1} + f = b k \Rightarrow g x_{1} + f y_{1} + c = c k$ .

So $x_{1} = a k - g$ and $y_{1} = b k - f$ .

Substitute into the third equation: $g (a k - g) + f (b k - f) + c = c k$ .

So $a g k - g^{2} + b f k - f^{2} + c = c k$ . Thus, $k (a g + b f - c) = g^{2} + f^{2} - c$ .

Hence, $k = \frac{g^{2} + f^{2} - c}{a g + b f - c}$ .

So $x_{1} = a \frac{g^{2} + f^{2} - c}{a g + b f - c} - g$ and $y_{1} = b \frac{g^{2} + f^{2} - c}{a g + b f - c} - f$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Its center is $(- g, - f)$ .

Let the point be $P (x_{1}, y_{1})$ . The polar of $P$ is

$x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$ .

So $(x_{1} + g) x + (y_{1} + f) y + (g x_{1} + f y_{1} + c) = 0$ .

Thus, the normal vector to the polar is $(x_{1} + g, y_{1} + f)$ .

Now consider the line joining the center $C (- g, - f)$ and the point $P (x_{1}, y_{1})$ .

Its direction vector is $(x_{1} + g, y_{1} + f)$ .

Hence, the polar is perpendicular to the line joining the point and the center.

The family of circles is $x^{2} + y^{2} + 2 p x + c = 0$ . Let the given point be $(x_{1}, y_{1})$ .

The polar of $(x_{1}, y_{1})$ with respect to the circle is $x x_{1} + y y_{1} + p (x + x_{1}) + c = 0$ .

So $(x_{1} + p) x + y_{1} y + p x_{1} + c = 0 \Rightarrow x_{1} x + y_{1} y + c + p (x + x_{1}) = 0$ .

Now observe that if $x + x_{1} = 0$ , the term containing $p$ vanishes.

So the equation reduces to $x_{1} x + y_{1} y + c = 0$ . Substitute $x = - x_{1}$ .

Then $- x_{1}^{2} + y_{1} y + c = 0$ . So $y = \frac{x_{1}^{2} - c}{y_{1}}$ .

Thus, the point $(- x_{1}, \frac{x_{1}^{2} - c}{y_{1}})$ satisfies the polar for all values of $p$ .

Hence, all polars pass through this fixed point.

The polar of $(α, β)$ with respect to $x^{2} + y^{2} = a^{2}$ is $α x + β y = a^{2}$ .

For this line to touch the circle $(x - a)^{2} + y^{2} = a^{2}$ , the distance from its center $(a, 0)$ to the line must equal the radius $a$ .

So $\frac{| α a - a^{2} |}{\sqrt{α^{2} + β^{2}}} = a \Rightarrow {(α a - a^{2})}^{2} = a^{2} (α^{2} + β^{2})$ .

$\Rightarrow (α - a)^{2} = α^{2} + β^{2} \Rightarrow β^{2} + 2 a α = a^{2}$ .

Thus, $(α, β)$ lies on the curve $y^{2} + 2 a x = a^{2}$ .

For the circle $x^{2} + y^{2} = a^{2}$ , the polar of $(x_{i}, y_{i})$ is $x x_{i} + y y_{i} = a^{2}$ .

So the polars of $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ are

$x x_{1} + y y_{1} = a^{2}, x x_{2} + y y_{2} = a^{2}$ , and $x x_{3} + y y_{3} = a^{2}$ .

These three lines are concurrent if there exists a point $(h, k)$ satisfying all three.

So $h x_{1} + k y_{1} = a^{2}, h x_{2} + k y_{2} = a^{2}$ , and $h x_{3} + k y_{3} = a^{2}$ .

Subtract pairwise: $h (x_{1} - x_{2}) + k (y_{1} - y_{2}) = 0$ and $h (x_{2} - x_{3}) + k (y_{2} - y_{3}) = 0$ .

For non-zero $(h, k)$ , these two equations imply $(x_{1} - x_{2}) (y_{2} - y_{3}) = (x_{2} - x_{3}) (y_{1} - y_{2})$ .

This is equivalent to $x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0$ .

Thus the points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ are collinear.

The given circles are $x^{2} + y^{2} - 2 x - 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 4 y - 6 = 0$ .

Compare each with the general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

For the first circle, we get $g_{1} = - 1, f_{1} = - 3$ , and $c_{1} = - 12$ .

For the second circle, we get $g_{2} = 3, f_{2} = 2$ , and $c_{2} = - 6$ .

Two circles cut orthogonally if $2 (g_{1} g_{2} + f_{1} f_{2}) = c_{1} + c_{2}$ .

Substituting the values $2 ((- 1) (3) + (- 3) (2)) = 2 (- 3 - 6) = - 18$ .

Also $c_{1} + c_{2} = - 12 - 6 = - 18$ .

Both sides are equal. Hence, the given circles cut each other orthogonally.

Let the two circles be $S = x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and $S_{1} = x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ .

Their radii are $a$ and $a_{1}$ . So $g^{2} + f^{2} - c = a^{2}$ and $g_{1}^{2} + f_{1}^{2} - c_{1} = a_{1}^{2}$ .

Consider the circle $\frac{S}{a} + \frac{S_{1}}{a_{1}} = 0$ .

Its equation is $(\frac{1}{a} + \frac{1}{a_{1}}) (x^{2} + y^{2}) + 2 (\frac{g}{a} + \frac{g_{1}}{a_{1}}) x + 2 (\frac{f}{a} + \frac{f_{1}}{a_{1}}) y + (\frac{c}{a} + \frac{c_{1}}{a_{1}}) = 0$ .

$= x^{2} + y^{2} + 2 G x + 2 F y + C = 0$ , where $G = \frac{\frac{g}{a} + \frac{g_{1}}{a_{1}}}{\frac{1}{a} + \frac{1}{a_{1}}}, F = \frac{\frac{f}{a} + \frac{f_{1}}{a_{1}}}{\frac{1}{a} + \frac{1}{a_{1}}}$ , and $C = \frac{\frac{c}{a} + \frac{c_{1}}{a_{1}}}{\frac{1}{a} + \frac{1}{a_{1}}}$ .

Now consider the circle $\frac{S}{a} - \frac{S_{1}}{a_{1}} = 0$ . Similarly it becomes $x^{2} + y^{2} + 2 G_{1} x + 2 F_{1} y + C_{1} = 0$

where $G_{1} = \frac{\frac{g}{a} - \frac{g_{1}}{a_{1}}}{\frac{1}{a} - \frac{1}{a_{1}}}, F_{1} = \frac{\frac{f}{a} - \frac{f_{1}}{a_{1}}}{\frac{1}{a} - \frac{1}{a_{1}}}$ , and $C_{1} = \frac{\frac{c}{a} - \frac{c_{1}}{a_{1}}}{\frac{1}{a} - \frac{1}{a_{1}}}$ .

Two circles cut orthogonally if $2 (G G_{1} + F F_{1}) = C + C_{1}$ . After simplification, both sides reduce to the same value.

Hence, the circles represented by $\frac{S}{a} \pm \frac{S_{1}}{a_{1}} = 0$ intersect at right angles.

Let the required circles pass through the points $(0, 0)$ and $(0, - a)$ .

Then their equation can be taken as $x^{2} + y^{2} + 2 g x + a y = 0$ since substituting $(0, 0)$ and $(0, - a)$ satisfies it.

Now this circle touches the line $y = m x + c$ . So the distance of the center $(- g, - \frac{a}{2})$ from the line is equal to the radius.

The radius is $\sqrt{g^{2} + \frac{a^{2}}{4}}$ . So $\frac{| - m g + \frac{a}{2} - c |}{\sqrt{m^{2} + 1}} = \sqrt{g^{2} + \frac{a^{2}}{4}}$ .

$\Rightarrow {(- m g + \frac{a}{2} - c)}^{2} = (m^{2} + 1) (g^{2} + \frac{a^{2}}{4})$ . This gives a quadratic in $g$ .

The two circles correspond to the two values of $g$ . Let them be $g_{1}$ and $g_{2}$ .

For the two circles to cut orthogonally, the condition is $2 (g_{1} g_{2} + \frac{a^{2}}{4}) = 0$ .

So $g_{1} g_{2} = - \frac{a^{2}}{4}$ .

From the quadratic equation in $g$ , the product of roots is ${(\frac{a}{2} - c)}^{2} - (m^{2} + 1) \frac{a^{2}}{4}$ all divided by $m^{2}$ .

So $g_{1} g_{2} = \frac{{(\frac{a}{2} - c)}^{2} - (m^{2} + 1)}{\frac{a^{2}}{4}} m^{2}$ . Equate this to $- \frac{a^{2}}{4}$ .

Hence, $c^{2} = a^{2} (2 + m^{2})$ .

Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Given circle one is $x^{2} + y^{2} + 3 x - 5 y + 6 = 0$ .

So $g_{1} = \frac{3}{2}, f_{1} = - \frac{5}{2}, c_{1} = 6$ . Given circle two is $4 x^{2} + 4 y^{2} - 28 x + 29 = 0 => x^{2} + y^{2} - 7 x + 29 / 4 = 0$ .

So $g_{2} = - \frac{7}{2}, f_{2} = 0, c_{2} = \frac{29}{4}$ .

For orthogonality with first circle $2 (g g_{1} + f f_{1}) = c + c_{1} \Rightarrow 3 g - 5 f = c + 6$ .

For orthogonality with second circle $2 (g g_{2} + f f_{2}) = c + c_{2} \Rightarrow c = - 7 g - \frac{29}{4}$ .

Substitute in first equation $3 g - 5 f = - 7 g - \frac{29}{4} + 6 \Rightarrow f = 2 g + \frac{1}{4}$ .

Now the center lies on the line $3 x + 4 y + 1 = 0$ . So $3 (- g) + 4 (- f) + 1 = 0$ .

Thus, $3 g + 4 f = 1$ . Substitute $f = 2 g + \frac{1}{4} \Rightarrow 3 g + 4 (2 g + \frac{1}{4}) = 1 \Rightarrow 11 g = 0$ so $g = 0$ .

Then, $f = \frac{1}{4}$ . Now $c = - 7 g - \frac{29}{4} = - \frac{29}{4}$ .

Thus, the required circle is $x^{2} + y^{2} + \frac{1}{2} y - \frac{29}{4} = 0$ .

Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Since it cuts the circle $x^{2} + y^{2} = 4$ orthogonally, the condition is $c + (- 4) = 0$ .

So $c = 4$ . Thus the circle becomes $x^{2} + y^{2} + 2 g x + 2 f y + 4 = 0$ . Its center is $(- g, - f)$ .

Given that the center lies on the line $2 x - 2 y + 9 = 0$ , so $2 (- g) - 2 (- f) + 9 = 0$ . Thus, $- 2 g + 2 f + 9 = 0$ or $f = g - \frac{9}{2}$ .

Substitute into the equation of the circle $x^{2} + y^{2} + 2 g x + 2 (g - \frac{9}{2}) y + 4 = 0$ .

So $x^{2} + y^{2} + 2 g (x + y) - 9 y + 4 = 0$ . Rewrite as $x^{2} + y^{2} - 9 y + 4 + 2 g (x + y) = 0$ .

This represents a family of circles depending on $g$ .

For fixed points, eliminate $g$ . So the condition is $x + y = 0$ .

Substitute $y = - x$ into the equation $x^{2} + x^{2} - 9 (- x) + 4 = 0$ .

So $2 x^{2} + 9 x + 4 = 0$ . $x = - \frac{1}{2}$ or $x = - 4$ . Thus, the corresponding $y$ values are $\frac{1}{2}$ and $4$ .

Hence, the two fixed points are $(- \frac{1}{2}, \frac{1}{2})$ and $(- 4, 4)$ .

Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Since it cuts the given circles orthogonally,

therefore $2 g g_{1} + 2 f f_{2} - c - c_{1} = 0 \dots$ (1) and $2 g g_{2} + 2 f f_{2} - c - c_{2} = 0 \dots$ (2)

Eliminating $g$ and $f$ from these equations gives us $| \begin{matrix} x^{2} + y^{2} + c & x & y \\ - c - c_{1} & g_{1} & f_{1} \\ - c - c_{2} & g_{2} & f_{2} \end{matrix} | = 0$

$| \begin{matrix} x^{2} + y^{2} & - x & - y \\ - c_{1} & g_{1} & f_{1} \\ - c_{2} & g_{2} & f_{2} \end{matrix} | + c | \begin{matrix} 1 & - x & - y \\ 1 & g_{1} & f_{1} \\ 1 & g_{2} & f_{2} \end{matrix} | = 0$ , which is of the form

$\Rightarrow | \begin{matrix} x^{2} + y^{2} & - x & - y \\ c_{1} & g_{1} & f_{1} \\ c_{2} & g_{2} & f_{2} \end{matrix} | + k | \begin{matrix} 1 & - x & - y \\ 1 & g_{1} & f_{1} \\ 1 & g_{2} & f_{2} \end{matrix} | = 0$

The given circles are $x^{2} + y^{2} + 5 x + 3 y + 7 = 0$ and $x^{2} + y^{2} - 8 x + 6 y + k = 0$ .

Compare them with the general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

For the first circle, $g_{1} = \frac{5}{2}, f_{1} = \frac{3}{2}$ , and $c_{1} = 7$ .

For the second circle, $g_{2} = - 4, f_{2} = 3$ , and $c_{2} = k$ .

Two circles cut orthogonally if $2 (g_{1} g_{2} + f_{1} f_{2}) = c_{1} + c_{2}$ .

Substitute the values, $2. \frac{5}{2} . (- 4) + \frac{3}{2} .3 = 7 + k$ .

Hence, $- 11 = 7 + k \Rightarrow k = - 18$ .

Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Since it passes through the origin, $c = 0$ .

Now consider the circle $x^{2} + y^{2} - 4 x + 6 y + 10 = 0$ .

Comparing with the general form, $g_{1} = - 2, f_{1} = 3$ , and $c_{1} = 10$ .

Since the circles cut orthogonally, $2 (g g_{1} + f f_{1}) = c + c_{1}$ .

Thus, $2 (- 2 g + 3 f) = 10 \Rightarrow - 2 g + 3 f = 5$ .

Now consider the second circle $x^{2} + y^{2} + 12 y + 6 = 0$ .

Comparing with the general form, $g_{2} = 0, f_{2} = 6$ , and $c_{2} = 6$ .

Again using the orthogonality condition, $2 (g g_{2} + f f_{2}) = c + c_{2} \Rightarrow f = 1 / 2$ .

Substitute into $- 2 g + 3 f = 5$ . So $- 2 g + \frac{3}{2} = 5 \Rightarrow g = - \frac{7}{4}$ .

Thus, the required circle is $x^{2} + y^{2} - \frac{7}{2} x + y = 0$ .

Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Since it passes through the origin, $c = 0$ .

Given that the center lies on the line $x + y + 4 = 0$ , so $- g - f + 4 = 0$ .

Now consider the circle $x^{2} + y^{2} - 4 x + 2 y + 4 = 0$ . Comparing with the general form, $g_{1} = - 2, f_{1} = 1$ , and $c_{1} = 4$ .

Since the circles cut orthogonally, $2 (g g_{1} + f f_{1}) = c + c_{1}$ . So $2 (- 2 g + f) = 4$ . Hence, $- 2 g + f = 2$ .

Solving the two equation gives $g = \frac{2}{3}$ and $f = 4 - \frac{2}{3} = \frac{10}{3}$ .

Thus, the required circle is $x^{2} + y^{2} + \frac{4}{3} x + \frac{20}{3} y = 0$ .

Let the third circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Let the other two circles be $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ and $x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$ .

Since the first two circles cut the third circle orthogonally, $2 (g g_{1} + f f_{1}) = c + c_{1}$ and $2 (g g_{2} + f f_{2}) = c + c_{2}$ .

The common chord of the first two circles is obtained by subtracting their equations.

So its equation is $2 (g_{1} - g_{2}) x + 2 (f_{1} - f_{2}) y + c_{1} - c_{2} = 0$ .

From orthogonality $\Rightarrow c_{1} - c_{2} = 2 g (g_{1} - g_{2}) + 2 f (f_{1} - f_{2})$ .

Substitute into the equation of the common chord. Then $2 (g_{1} - g_{2}) x + 2 (f_{1} - f_{2}) y + 2 g (g_{1} - g_{2}) + 2 f (f_{1} - f_{2}) = 0$ .

Factor, $(g_{1} - g_{2}) (x + g) + (f_{1} - f_{2}) (y + f) = 0$ . The center of the third circle is $(- g, - f)$ .

Substitute $x = - g$ and $y = - f$ . The equation is satisfied.

Hence, the common chord passes through the center of the third circle.

Let the required circle be $S = x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Let $S_{1} = x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0, S_{2} = x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$ and $S_{3} = x^{2} + y^{2} + 2 g_{3} x + 2 f_{3} y + c_{3} = 0$ .

Since the circle $S = 0$ cuts each of these orthogonally,

$2 (g g_{1} + f f_{1}) = c + c_{1}, 2 (g g_{2} + f f_{2}) = c + c_{2}$ , and $2 (g g_{3} + f f_{3}) = c + c_{3}$ .

Now consider the circle $k S_{1} + l S_{2} + m S_{3} = 0$ .

Its equation is $(k + l + m) (x^{2} + y^{2}) + 2 (k g_{1} + l g_{2} + m g_{3}) x + 2 (k f_{1} + l f_{2} + m f_{3}) y + (k c_{1} + l c_{2} + m c_{3}) = 0$ .

Divide throughout by $(k + l + m)$ . Then the circle becomes $x^{2} + y^{2} + 2 G x + 2 F y + C = 0$ , where $G = \frac{k g_{1} + l g_{2} + m g_{3}}{k + l + m}, F = \frac{k f_{1} + l f_{2} + m f_{3}}{k + l + m}$ , and $C = \frac{k c_{1} + l c_{2} + m c_{3}}{k + l + m}$ .

$2 (g G + f F) = \frac{k (2 (g g_{1} + f f_{1})) + l (2 (g g_{2} + f f_{2})) + m (2 (g g_{3} + f f_{3}))}{k + l + m}$ .

Using the orthogonality conditions, $2 (g G + f F) = \frac{k (c + c_{1}) + l (c + c_{2}) + m (c + c_{3})}{k + l + m}$ .

$2 (g G + f F) = c + \frac{k c_{1} + l c_{2} + m c_{3}}{k + l + m} = c + C$ .

Hence, the circle $k S_{1} + l S_{2} + m S_{3} = 0$ cuts the circle $S = 0$ orthogonally.

Any circle passing through $(0, k)$ and $(0, - k)$ has equation $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Since $(0, k)$ lies on it, $k^{2} + 2 f k + c = 0$ . Since $(0, - k)$ lies on it, $k^{2} - 2 f k + c = 0$ .

Subtracting, $4 f k = 0$ . So $f = 0$ . Then $c = - k^{2}$ . Hence, the circle is $x^{2} + y^{2} + 2 g x - k^{2} = 0$ .

Its center is $(- g, 0)$ and radius is $\sqrt{g^{2} + k^{2}}$ .

Now the circle touches the line $y = m x + b$

Therefore, the perpendicular distance of the center from the line equals the radius.

So $\frac{| - m g + b |}{\sqrt{m^{2} + 1}} = \sqrt{g^{2} + k^{2}} \Rightarrow b^{2} - 2 b m g + m^{2} g^{2} = m^{2} g^{2} + g^{2} + k^{2} m^{2} + k^{2}$ .

Thus, $g^{2} + 2 b m g + k^{2} (m^{2} + 1) - b^{2} = 0$ . This quadratic gives the two possible circles.

Let their corresponding parameters be $g_{1}$ and $g_{2}$ . Then $g_{1} g_{2} = k^{2} (m^{2} + 1) - b^{2}$ .

Now the two circles are $x^{2} + y^{2} + 2 g_{1} x - k^{2} = 0$ and $x^{2} + y^{2} + 2 g_{2} x - k^{2} = 0$ .

They cut orthogonally if $2 (g_{1} g_{2}) = - 2 k^{2}$ .

So $g_{1} g_{2} = - k^{2}$ . Hence, $k^{2} (m^{2} + 1) - b^{2} = - k^{2}$ . Therefore, $b^{2} = k^{2} (m^{2} + 2)$ .

Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + k = 0$ .

Since it cuts the circle $x^{2} + y^{2} = c^{2}$ orthogonally, the condition is $2 (g * 0 + f * 0) = k - c^{2}$ .

So $k = - c^{2}$ . Hence the general equation of the circle is $x^{2} + y^{2} + 2 g x + 2 f y - c^{2} = 0$ .

Now suppose it passes through the point $(a, b)$ .

Substituting, $a^{2} + b^{2} + 2 g a + 2 f b - c^{2} = 0$ . So $2 g a + 2 f b = c^{2} - a^{2} - b^{2}$ .

Now consider the point $(\frac{c^{2} a}{a^{2} + b^{2}}, \frac{c^{2} b}{a^{2} + b^{2}})$ .

Substitute this point into the equation of the circle. We get

$\frac{c^{4} a^{2}}{{(a^{2} + b^{2})}^{2}} + \frac{c^{4} b^{2}}{{(a^{2} + b^{2})}^{2}} + \frac{2 g c^{2} a}{(a^{2} + b^{2})} + \frac{2 f c^{2} b}{(a^{2} + b^{2})} - c^{2}$ .

Combine the first two terms, $= \frac{c^{4} (a^{2} + b^{2})}{{(a^{2} + b^{2})}^{2}} + \frac{2 c^{2} (g a + f b)}{a^{2} + b^{2}} - c^{2}$ .

$\Rightarrow \frac{c^{4}}{a^{2} + b^{2}} + \frac{2 c^{2} (g a + f b)}{a^{2} + b^{2}} - c^{2}$ .

Using $2 g a + 2 f b = c^{2} - a^{2} - b^{2}$ , we get $2 (g a + f b) = c^{2} - a^{2} - b^{2}$ .

Substitute, $= \frac{c^{4}}{a^{2} + b^{2}} + \frac{c^{2} (c^{2} - a^{2} - b^{2})}{a^{2} + b^{2}} - c^{2}$ .

Simplify, $= \frac{c^{4} + c^{4} - c^{2} (a^{2} + b^{2})}{a^{2} + b^{2}} - c^{2} = \frac{2 c^{4} - c^{2} (a^{2} + b^{2})}{a^{2} + b^{2}} - c^{2} = 0$ .

Hence, the circle also passes through $(\frac{c^{2} a}{a^{2} + b^{2}}, \frac{c^{2} b}{a^{2} + b^{2}})$ .

Let the given circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Let $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ be conjugate points with respect to this circle.

Therefore the polar of $P$ passes through $Q$ . So $x_{1} x_{2} + y_{1} y_{2} + g (x_{1} + x_{2}) + f (y_{1} + y_{2}) + c = 0$ .

Now consider the circle having $P Q$ as diameter. Its equation is $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$ .

Comparing with $x^{2} + y^{2} + 2 G x + 2 F y + C = 0$ , we get $G = - \frac{x_{1} + x_{2}}{2}, F = - \frac{y_{1} + y_{2}}{2}$ and $C = x_{1} x_{2} + y_{1} y_{2}$ .

For orthogonality with the given circle, the condition is $2 (g G + f F) = c + C$ .

Substitute the values, $2 (g G + f F) = 2 g (- \frac{x_{1} + x_{2}}{2}) + 2 f (- \frac{y_{1} + y_{2}}{2}) = - g (x_{1} + x_{2}) - f (y_{1} + y_{2})$ .

Also $c + C = c + x_{1} x_{2} + y_{1} y_{2}$ .

Using the conjugate point relation, $x_{1} x_{2} + y_{1} y_{2} + g (x_{1} + x_{2}) + f (y_{1} + y_{2}) + c = 0$ ,

we get $c + x_{1} x_{2} + y_{1} y_{2} = - g (x_{1} + x_{2}) - f (y_{1} + y_{2})$ .

Hence, $2 (g G + f F) = c + C$ . Therefore, the circle on $P Q$ as diameter cuts the circle $S$ orthogonally.

Equation of common chord of the circle is $x + y - 7 = 0$ , which is also the radical axis.

Center of second circle is $(4, 3)$ , which lies on the line obtained. Hence, the line is a diamter of the second circle, and hence, the circumference of the second circle is bisected.

The given circles are $2 x^{2} + 2 y^{2} - 2 x + 6 y - 3 = 0$ and $x^{2} + y^{2} + 4 x + 2 y + 1 = 0$ .

Divide the first equation by $2$ . Then the circles become $S_{1} = x^{2} + y^{2} - x + 3 y - \frac{3}{2} = 0$ and $S_{2} = x^{2} + y^{2} + 4 x + 2 y + 1 = 0$ .

Any circle coaxal with them is $S_{1} + λ S_{2} = 0$ . So $(1 + λ) (x^{2} + y^{2}) + (- 1 + 4 λ) x + (3 + 2 λ) y + (- \frac{3}{2} + λ) = 0$ .

Divide throughout by $(1 + λ)$ . Then the circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , where $2 g = \frac{- 1 + 4 λ}{1 + λ}$ and $2 f = \frac{3 + 2 λ}{1 + λ}$ .

Hence, the center is $(\frac{- 4 λ + 1}{2 (1 + λ)}, - \frac{3 + 2 λ}{2 (1 + λ)})$ .

Subtracting, $5 x - y + \frac{5}{2} = 0$ . So the center must satisfy $5 \frac{- 4 λ + 1}{2 (1 + λ)} - \frac{- (3 + 2 λ)}{2 (1 + λ)} + \frac{5}{2} = 0$ .

$\Rightarrow 5 (- 4 λ + 1) + (3 + 2 λ) + 5 (1 + λ) = 0 \Rightarrow - 20 λ + 5 + 3 + 2 λ + 5 + 5 λ = 0$ .

$\Rightarrow 13 - 13 λ = 0 \Rightarrow λ = 1$ .

Therefore, the required circle is $S_{1} + S_{2} = 0$ . So $2 x^{2} + 2 y^{2} + 3 x + 5 y - \frac{1}{2} = 0$ .

The given circles are $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and $2 x^{2} + 2 y^{2} + 3 x + 8 y + 2 c = 0$ .

The radical axis is obtained by subtracting the equations. So $(2 g - \frac{3}{2}) x + (2 f - 4) y = 0$ .

Hence, the radical axis is $(4 g - 3) x + (4 f - 8) y = 0$ .

Now this line touches the circle $x^{2} + y^{2} + 2 x - 2 y + 1 = 0$ . Its center is $(- 1, 1)$ and radius is $\sqrt{1 - 1 + 1} = 1$ .

Therefore the perpendicular distance from the center to the line equals the radius.

So $\frac{| (4 g - 3) (- 1) + (4 f - 8) (1) |}{\sqrt{(4 g - 3)^{2} + (4 f - 8)^{2}}} = 1$ .

$\Rightarrow (4 g - 3) (f - 2) = 0$ . Therefore, either $g = \frac{3}{4}$ or $f = 2$ .

The given circles are $x^{2} + y^{2} + 2 x + 4 y - 6 = 0$ and $x^{2} + y^{2} = 4$ .

Their radical axis is obtained by subtraction. So $(x^{2} + y^{2} + 2 x + 4 y - 6) - (x^{2} + y^{2} - 4) = 0$ .

Hence, the radical axis is $x + 2 y - 1 = 0$ .

Let one circle of the required family be $x^{2} + y^{2} - 4 = 0$ .

Then every circle having the same radical axis with it is obtained by adding a multiple of the radical axis.

Hence, the required family is $x^{2} + y^{2} - 4 + λ (x + 2 y - 1) = 0$ .

Therefore, the general equation is $x^{2} + y^{2} + λ x + 2 λ y - (λ + 4) = 0$ , where $λ$ is an arbitrary parameter.

Let the required point be $(h, k)$ . The square of the length of the tangent from $(h, k)$ to a circle is obtained by substituting the point in the equation of the circle.

For the circle $x^{2} + y^{2} = 1$ , the tangent length squared is $h^{2} + k^{2} - 1$ .

For the circle $x^{2} + y^{2} - 8 x + 15 = 0$ , the tangent length squared is $h^{2} + k^{2} - 8 h + 15$ .

Since the tangent lengths are equal, $h^{2} + k^{2} - 1 = h^{2} + k^{2} - 8 h + 15$ .

Thus, $h = 2$ . Now consider the third circle $x^{2} + y^{2} + 10 y + 24 = 0$ .

The tangent length squared is $h^{2} + k^{2} + 10 k + 24$ .

Again equating tangent lengths, $h^{2} + k^{2} - 1 = h^{2} + k^{2} + 10 k + 24 \Rightarrow k = - \frac{5}{2}$ .

Therefore, the required point is $(2, - \frac{5}{2})$ .

Let the two circles be $S_{1} = x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ and $S_{2} = x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$ .

Let the given point be $P (x_{1}, y_{1})$ .

The polar of $P$ with respect to $S_{1}$ is $x x_{1} + y y_{1} + g_{1} (x + x_{1}) + f_{1} (y + y_{1}) + c_{1} = 0$ .

The polar of $P$ with respect to $S_{2}$ is $x x_{1} + y y_{1} + g_{2} (x + x_{1}) + f_{2} (y + y_{1}) + c_{2} = 0$ .

These two polars meet at the point $Q$ . Subtract the two equations.

Then the coordinates of $Q$ satisfies $(g_{1} - g_{2}) x + (f_{1} - f_{2}) y + g_{1} x_{1} - g_{2} x_{1} + f_{1} y_{1} - f_{2} y_{1} + c_{1} - c_{2} = 0$ .

So $(g_{1} - g_{2}) (x + x_{1}) + (f_{1} - f_{2}) (y + y_{1}) + c_{1} - c_{2} = 0$ .

Now the radical axis of the two circles is $2 (g_{1} - g_{2}) x + 2 (f_{1} - f_{2}) y + c_{1} - c_{2} = 0$ .

Let the midpoint of $P Q$ be $(h, k)$ . Then $h = \frac{x + x_{1}}{2}$ and $k = \frac{y + y_{1}}{2}$ .

Substitute in the equation obtained above.

We get $2 (g_{1} - g_{2}) h + 2 (f_{1} - f_{2}) k + c_{1} - c_{2} = 0$ . But this is exactly the equation of the radical axis. Hence, the midpoint of $P Q$ lies on the radical axis.

Let the given points be $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ .

Let $P (x, y)$ be a point such that $\frac{P A}{P B} = k$ , where $k$ is a constant and $k \neq 1$ .

Then $\frac{\sqrt{(x - x_{1})^{2} + (y - y_{1})^{2}}}{\sqrt{(x - x_{2})^{2} + (y - y_{2})^{2}}} = k$ .

Squaring, $(x - x_{1})^{2} + (y - y_{1})^{2} = k^{2} ((x - x_{2})^{2} + (y - y_{2})^{2})$ .

Expand both sides, $x^{2} - 2 x x_{1} + x_{1}^{2} + y^{2} - 2 y y_{1} + y_{1}^{2} = k^{2} (x^{2} - 2 x x_{2} + x_{2}^{2} + y^{2} - 2 y y_{2} + y_{2}^{2})$ .

$\Rightarrow (1 - k^{2}) (x^{2} + y^{2}) + 2 (k^{2} x_{2} - x_{1}) x + 2 (k^{2} y_{2} - y_{1}) y + x_{1}^{2} + y_{1}^{2} - k^{2} (x_{2}^{2} + y_{2}^{2}) = 0$ .

Since $k \neq 1$ , divide by $(1 - k^{2})$ . The equation becomes of the form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Hence, the locus is a circle.

At $A, P A = 0$ . So the condition $\frac{P A}{P B} = k$ gives $0 = k$ , which is impossible since $k$ is a fixed non-zero constant.

Hence, the circle does not pass through $A$ .

Similarly, at $B, P B = 0$ , so the ratio becomes infinite, which is impossible.

Hence, the circle does not pass through $B$ .

Let $A B$ and $C D$ be the two rods of lengths $a$ and $b$ respectively.

Let the equation of the circle passing through points $A, B, C$ and $D$ be $x^{2} + y^{2} - 2 α x - 2 β y + λ = 0$ , whose center is $P (α, β)$ .

Putting $y = 0$ gives us $x^{2} - 2 α x + λ = 0 \Rightarrow x_{1} + x_{2} = 2 α, x_{1} x_{2} = λ$

$∵ a = | x_{1} - x_{2} |, ∴ a^{2} = (x_{1} - x_{2})^{2} = (x_{1} + x_{2})^{2} - 4 x_{1} x_{2} = 4 (α^{2} - λ)$

Similarly, $b^{2} = 4 (β^{2} - λ) ∴ a^{2} - b^{2} = 4 (α^{2} - β^{2})$ .

Hence, the locus of the point $P (α, β)$ is $4 (x^{2} - y^{2}) = a^{2} - b^{2}$ .

Let $A$ and $B$ be two fixed points. Let $A B = 2 a$ , we take the mid-point $O$ of $A B$ as the origin and $O B$ as $x$ -axis. Let $A = (- a, 0)$ and $B = (a, 0)$ .

Let one straight line which rotates about $B$ makes an angle $t h e t a$ with the $x$ -axis at any time $t$ and at that time the second line which rotates about $A$ makes an angle $2 θ$ with $x$ -axis.

Now equations of these lines are $y = \tan θ (x - a) \dots$ (1) and $y = \tan 2 θ (x + a)$

Solving we get $x = \frac{a (\tan 2 θ + t a n θ)}{\tan θ - \tan 2 θ}$ and $y = \frac{2 a . (\tan 2 θ \tan θ)}{\tan θ - \tan 2 θ}$

$\Rightarrow x + a = - 2 a \cos 2 θ$ and $y = - 2 a \sin 2 θ \Rightarrow (x + a)^{2} + y^{2} = 4 a^{2}$ is the required locus.

Let $O A = a, O B = b$ . The equation of the circle through $O, A, B$ is $x^{2} + y^{2} - a x - b y = 0$

Its radius is $r = \sqrt{{(- \frac{a}{2})}^{2} + {(- \frac{b}{2})}^{2}} \Rightarrow a^{2} + b^{2} = 4 r^{2}$ . Here, $a, b$ are variables and $r$ is a constant.

Let $P (α, β)$ and $∠ P O A = θ$ , then $α = O P \cos θ = r \cos θ$ and $β = r \sin θ$

Equation of line $A B$ is $\frac{x}{a} + \frac{y}{b} = 1. P$ lies on the line $A B \Rightarrow \frac{α}{a} + \frac{β}{b} = 1$ .

$∵ O P ⊥ A B ∴ \frac{β}{α} \frac{b - 0}{0 - a} = - 1 \Rightarrow a α = b β = k (let) \Rightarrow a = \frac{α}{k}, b = \frac{k}{β}$

$\Rightarrow \frac{α}{\frac{k}{α}} + \frac{β}{\frac{k}{β}} = 1 \Rightarrow α^{2} + β^{2} = k \Rightarrow a = \frac{α^{2} + β^{2}}{α}, b = \frac{α^{2} + β^{2}}{β}$

$\Rightarrow {(α^{2} + β^{2})}^{2} (\frac{1}{α^{2}} + \frac{1}{β^{2}}) = 4 r^{2}$

Thus, locus of point $P$ is ${(x^{2} + y^{2})}^{2} (\frac{1}{x^{2}} + \frac{1}{y^{2}}) = 4 r^{2}$ .

Let $P (α, β)$ be the point whose locus is to be found. Let the given circle be $x^{2} + y^{2} = a^{2}$ , and tangent to this circle is $y = m x + a \sqrt{1 + m^{2}}$ which passes through $P$ .

Thus, $β = m α + a \sqrt{1 + m^{2}} \Rightarrow m^{2} (a^{2} - α^{2}) + 2 m α β + a^{2} - β^{2} = 0$ .

This is a quadratic equation in $m$ and hance two values are possible. Thus, these lines will be orthogonal if $m_{1} m_{2} = - 1 \Rightarrow \frac{a^{2} - β^{2}}{a^{2} - α^{2}} = - 1 \Rightarrow α^{2} + β^{2} = 2 a^{2}$ .

Thus, locus of $P (α, β)$ is $x^{2} + y^{2} = 2 a^{2}$ .

Given is the parametric equation of the circle. The cartesian equation will be $x^{2} + y^{2} = a^{2}$ .

Let $A = (a \cos θ, a \sin θ), B = (a \cos \frac{θ + π}{3}, a \sin \frac{θ + π}{3})$ .

Equation of tangent at $A$ is $x a \cos θ + y a \sin θ = a^{2} \Rightarrow x \cos θ + y \sin θ = a$ .

Similarly, equation of tangent at $B$ is $x \cos \frac{θ + π}{3} + y \sin \frac{θ + π}{3} = a$

$\Rightarrow x \cos θ . \frac{1}{2} - x \sin θ \frac{\sqrt{3}}{2} + y \sin θ . \frac{1}{2} + y \cos θ \frac{\sqrt{3}}{2} = a$

$\Rightarrow \frac{1}{2} (x \cos θ + y \sin θ) + \frac{\sqrt{3}}{2} (y \cos θ - x \sin θ) = a$

$\Rightarrow \frac{a}{2} + \frac{\sqrt{3}}{2} (y \cos θ - x \sin θ) = a \Rightarrow y \cos θ - x \sin θ = \frac{a}{\sqrt{3}}$

Squaring and adding with the equation of tangent at $A$ yields

$3 (x^{2} + y^{2}) = 4 a^{2}$ , which is the required locus.

Equation of chord of intersection is $2 (a - b) x = 0 \Rightarrow x = 0$ .

Thus, $x = 0$ is the equation of the chord of intersection. $O A^{2} = \sqrt{a^{2} + c^{2} - a^{2}} = a$ .

Since common chord of the two circles is $y$ -axis and their centers are $(- a, 0)$ and $(- b, 0)$ lying on the $x$ -axis.

Therefore, one of $a q$ and $b$ will be positive and other negative. WLOG we can assume that $a < 0, b > 0$ with $| a | < b$ .

Let $A P$ be an arbitrary line through $A (0, c)$ which meets first circle at $P (x_{2}, y_{2})$ . Let the slope of $A P$ be $m$ .

Equation of $A P$ is $y = m x + c$ and that of $B Q$ is $y = m x - c$ .

Let $Q = (x_{3}, y_{3})$ . Let $R (α, β)$ be the mid-point of $P Q$ . Putting $y = m x + c$ in first circle yields

$x^{2} + (m x + c)^{2} + 2 a x - c^{2} = 0 \Rightarrow x = 0, - \frac{2 (a + c m)}{1 + m^{2}}$

$∵ x_{2} \neq 0 ∴ x_{2} = - \frac{2 (a + c m)}{1 + m^{2}}$ and $y_{2} = - \frac{2 m (a + c m)}{1 + m^{2}} + c$

Replacing $a$ by $b$ and $c$ by $- c$ gives us

$x_{3} = \frac{2 (b - c m)}{1 + m^{2}}$ and $y_{3} = - \frac{2 m (b - c m)}{1 + m^{2}} - c$

$α = f r a c x_{2} + x_{3} 2 = - \frac{a + b}{1 + m^{2}}, β = - \frac{m (a + b)}{1 + m^{2}} \Rightarrow \frac{β}{α} = m$ .

$\Rightarrow α = - \frac{(a + b) α^{2})}{α^{2} + β^{2}} \Rightarrow α^{2} + β^{2} + (a + b) α = 0$

Hence, the locus of $(α, β)$ is $x^{2} + y^{2} + (a + b) x = 0$ .

The chord of contact of tangents drawn from $(α, β)$ to the circle $x^{2} + y^{2} = a^{2}$ is $α x + β y = a^{2}$ .

This chord subtends a right angle at the center $(0, 0)$ . Let the perpendicular distance from the center to the chord be $d$ .

For a chord of a circle of radius $a$ subtending a right angle at the center, $d = a \cos \frac{π}{4} = \frac{a}{\sqrt{2}}$ .

Now the distance of the center from the chord $α x + β y - a^{2} = 0$ is $\frac{a^{2}}{\sqrt{α^{2} + β^{2}}}$ .

Therefore, $\frac{a^{2}}{\sqrt{α^{2} + β^{2}}} = \frac{a}{\sqrt{2}}$ . So $a^{4} = a^{2} \frac{α^{2} + β^{2}}{2}$ .

Hence, $α^{2} + β^{2} = 2 a^{2}$ . Therefore, the required condition is $α^{2} + β^{2} = 2 a^{2}$ .

Thus, the locus of $(α, β)$ is the circle $x^{2} + y^{2} = 2 a^{2}$ .

Let the tangents to the circles $x^{2} + y^{2} = a^{2}$ and $x^{2} + y^{2} = b^{2}$ intersect at the point $(h, k)$ .

The tangent from $(h, k)$ to the first circle has equation $y = m x \pm \sqrt{a^{2} (1 + m^{2})}$ .

Since it passes through $(h, k), k = m h \pm \sqrt{a^{2} (1 + m^{2})} \Rightarrow (k - m h)^{2} = a^{2} (1 + m^{2})$ .

This gives the combined equation of tangents from $(h, k)$ to the circle $x^{2} + y^{2} = a^{2}$ ,

namely ${(x h + y k - a^{2})}^{2} = (h^{2} + k^{2} - a^{2}) (x^{2} + y^{2} - a^{2})$ .

Similarly, the pair of tangents from $(h, k)$ to the circle $x^{2} + y^{2} = b^{2}$ is

${(x h + y k - b^{2})}^{2} = (h^{2} + k^{2} - b^{2}) (x^{2} + y^{2} - b^{2})$ .

Now the two tangents are mutually perpendicular. For a pair of tangents drawn from a point to the circle $x^{2} + y^{2} = r^{2}$ , the angle between them is a right angle if $h^{2} + k^{2} = 2 r^{2}$ .

Applying this separately to the two circles and combining for perpendicular tangents, we obtain $h^{2} + k^{2} = a^{2} + b^{2}$ .

Hence, the locus of the point of intersection is $x^{2} + y^{2} = a^{2} + b^{2}$ . This is a circle concentric with the given circles.

Let $(h, k)$ be the point from which tangents are drawn to the circle $x^{2} + y^{2} = a^{2}$ .

Let the angle between the tangents be $a l p h a$ . If $P$ is the external point and $O$ is the center, then in the right triangle formed by joining the center to the point of contact,

$\sin \frac{α}{2} = \frac{a}{\sqrt{h^{2} + k^{2}}}$ . Hence, $h^{2} + k^{2} = a^{2} \csc^{2} \frac{α}{2}$ .

Now use the identity $\csc^{2} \frac{α}{2} = \frac{\tan^{2} α + 4}{4 \tan^{2} \frac{α}{2}}$ .

After simplification, ${(h^{2} + k^{2} - 2 a^{2})}^{2} \tan^{2} α = 4 a^{2} (h^{2} + k^{2} - a^{2})$ .

Replacing $(h, k)$ by the general point $(x, y)$ , the locus is ${(x^{2} + y^{2} - 2 a^{2})}^{2} \tan^{2} α = 4 a^{2} (x^{2} + y^{2} - a^{2})$ .

Let the variable line through the fixed point $(h, k)$ have slope $m$ . Its equation is $y - k = m (x - h)$ .

So $m x - y + (k - m h) = 0$ . Let $(x, y)$ be the foot of the perpendicular drawn from the origin to this line.

Since $(x, y)$ lies on the line, $m x - y + k - m h = 0$ .

Also the line joining the origin to $(x, y)$ is perpendicular to the given line.

The slope of the given line is $m$ . Hence, the slope of the perpendicular from the origin is $- \frac{1}{m}$ .

Therefore, $\frac{y}{x} = - \frac{1}{m}$ , so $m = - \frac{x}{y}$ .

Substitute this in the line equation, $(- \frac{x}{y}) x - y + k - (- \frac{x}{y}) h = 0 \Rightarrow - x^{2} - y^{2} + k y + h x = 0$ .

Hence, the locus is $x^{2} + y^{2} - h x - k y = 0$ .

Take the fixed point $O$ as the origin. Let the two fixed parallel lines be $x = a$ and $x = - b$ .

Then the points $A (a, 0)$ and $B (- b, 0)$ lie on the perpendicular through $O$ .

Let $P (a, p)$ and $Q (- b, q)$ . Since $∠ P O Q$ is a right angle, the slopes of $O P$ and $O Q$ satisfy

$\frac{p}{a} (\frac{q}{- b}) = - 1$ . So $p q = a b$ . Now find the equation of the line $P Q$ .

Using the two-point form, $y - p = \frac{(q - p) (x - a)}{- b - a}$ .

This simplifies to $(p - q) x + (a + b) y - (a q + b p) = 0$ .

Let $(h, k)$ be the foot of the perpendicular from the origin to this line.

Then $h = - \frac{(p - q) (- a q - b p)}{(p - q)^{2} + (a + b)^{2}}$ and $k = - \frac{(a + b) (- a q - b p)}{(p - q)^{2} + (a + b)^{2}}$ .

So $h = \frac{(p - q) (a q + b p)}{(p - q)^{2} + (a + b)^{2}}$ and $k = \frac{(a + b) (a q + b p)}{(p - q)^{2} + (a + b)^{2}}$ .

Using $p q = a b$ , simplification gives $h^{2} + (k^{2}) = (a - b) h$ .

Rewrite, $h^{2} - (a - b) h + k^{2} = 0$ . Complete the square, ${(h - \frac{a - b}{2})}^{2} + k^{2} = \frac{(a + b)^{2}}{4}$ .

This is the equation of the circle whose diameter has endpoints $(a, 0)$ and $(- b, 0)$ , that is, the circle on $A B$ as diameter.

Hence, the locus of the foot of the perpendicular from $O$ to $P Q$ is the circle on $A B$ as diameter.

Let the other end of the diameter through $P (1, 2)$ be $(x, y)$ . Let the center of the circle be $(h, k)$ .

Since the center is the midpoint of the diameter joining $(1, 2)$ and $(x, y), h = \frac{x + 1}{2}$ and $k = \frac{y + 2}{2}$ .

The circle touches the $x$ -axis. Therefore, the radius equals the distance of the center from the $x$ -axis.

So the radius is $k$ . Now the radius is also half the length of the diameter.

Hence, $\frac{(x - 1)^{2} + (y - 2)^{2}}{4} = k^{2}$ . Substitute $k = \frac{y + 2}{2}$ .

Then $\frac{(x - 1)^{2} + (y - 2)^{2}}{4} = \frac{(y + 2)^{2}}{4} \Rightarrow (x - 1)^{2} = 8 y$ .

Let the required point be $(x, y)$ . The length of the tangent from $(x, y)$ to the circle

$x^{2} + y^{2} = a^{2}$ is $\sqrt{x^{2} + y^{2} - a^{2}}$ .

Similarly, the length of the tangent from $(x, y)$ to the circle $x^{2} + y^{2} = b^{2}$ is $\sqrt{x^{2} + y^{2} - b^{2}}$ .

Given that the tangent lengths vary inversely as the radii, $\frac{\sqrt{x^{2} + y^{2} - a^{2}}}{\sqrt{x^{2} + y^{2} - b^{2}}} = \frac{b}{a}$ .

$\Rightarrow \frac{x^{2} + y^{2} - a^{2}}{x^{2} + y^{2} - b^{2}} = \frac{b^{2}}{a^{2}} \Rightarrow a^{2} (x^{2} + y^{2} - a^{2}) = b^{2} (x^{2} + y^{2} - b^{2})$ .

Hence, $x^{2} + y^{2} = a^{2} + b^{2}$ . Therefore, the locus is the circle $x^{2} + y^{2} = a^{2} + b^{2}$ .

Take the square with sides parallel to the axes and center at the origin.

Since the side of the square is unity, its sides are $x = \frac{1}{2}, x = - \frac{1}{2}, y = \frac{1}{2}$ , and $y = - \frac{1}{2}$ .

Let $(x, y)$ be the moving point. Its perpendicular distances from the four sides are $| x - \frac{1}{2} |, | x + \frac{1}{2} |, | y - \frac{1}{2} |$ , and $| y + \frac{1}{2} |$ .

Given that the sum of their squares is $9, {(x - \frac{1}{2})}^{2} + {(x + \frac{1}{2})}^{2} + {(y - \frac{1}{2})}^{2} + {(y + \frac{1}{2})}^{2} = 9$ .

$\Rightarrow x^{2} + y^{2} = 4$ . This is a circle centered at the origin, which is the center of the square.

Therefore the locus is a circle concentric with the square. Its radius is $2$ .

The given circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Its center is $(- g, - f)$ .

Let the center be $(h, k)$ . Then $g = - h$ and $f = - k$ .

The length of the tangent from the origin to the circle is $\sqrt{c}$ evaluated at the origin, so $\sqrt{0 + 0 + 0 + 0 + c} = \sqrt{c}$ .

The pair of tangents drawn from the origin are perpendicular.

For tangents from a point to a circle to be perpendicular, the point must lie on the director circle.

The director circle of $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ is $x^{2} + y^{2} + 2 g x + 2 f y + 2 c - g^{2} - f^{2} = 0$ .

Since the origin lies on it, $2 c - g^{2} - f^{2} = 0$ . So $g^{2} + f^{2} = 2 c$ .

Now the radius squared of the circle is $g^{2} + f^{2} - c$ . Substitute the above relation, $r^{2} = c$ .

Thus, $g^{2} + f^{2} = 2 c$ . Replacing $g = - h$ and $f = - k$ , $h^{2} + k^{2} = 2 c$ .

Hence, the locus of the center is $x^{2} + y^{2} = 2 c$ .

Let the required circle have center $(h, k)$ and radius $r$ . The circle $x^{2} + y^{2} = a^{2}$ has center $(0, 0)$ and radius $a$ .

The circle $x^{2} + y^{2} = 4 a x$ . So its center is $(2 a, 0)$ and radius is $2 a$ .

Since the required circle touches both circles externally, $\sqrt{h^{2} + k^{2}} = r + a$ , and $\sqrt{(h - 2 a)^{2} + k^{2}} = r + 2 a$ .

Subtracting, $\sqrt{(h - 2 a)^{2} + k^{2}} - \sqrt{h^{2} + k^{2}} = a$ .

Let $d_{1} = \sqrt{h^{2} + k^{2}}$ , and $d_{2} = \sqrt{(h - 2 a)^{2} + k^{2}}$ .

Then $d_{2} = d_{1} + a$ . Squaring, $(h - 2 a)^{2} + k^{2} = h^{2} + k^{2} + 2 a d_{1} + a^{2}$ .

Simplify, $h^{2} - 4 a h + 4 a^{2} + k^{2} = h^{2} + k^{2} + 2 a d_{1} + a^{2}$ .

So $- 4 a h + 3 a^{2} = 2 a d_{1}$ . Hence, $d_{1} = \frac{3 a - 4 h}{2}$ .

Now square again, $h^{2} + k^{2} = \frac{(3 a - 4 h)^{2}}{4}$ . Thus, $4 h^{2} + 4 k^{2} = 9 a^{2} - 24 a h + 16 h^{2}$ .

Therefore, $12 h^{2} - 4 k^{2} - 24 a h + 9 a^{2} = 0$ .

Replacing $(h, k)$ by $(x, y)$ , the locus is $12 x^{2} - 4 y^{2} - 24 a x + 9 a^{2} = 0$ .

Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Its center is $(- g, - f)$ .

The given circles are $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$ , and $x^{2} + y^{2} - 4 x + 6 y + 4 = 0$ .

For the first circle, $g_{1} = 2, f_{1} = - 3$ , and $c_{1} = 9$ .

Since the required circle cuts it orthogonally, $2 (g g_{1} + f f_{1}) = c + c_{1}$ .

Thus, $4 g - 6 f = c + 9$ . For the second circle, $g_{2} = - 2, f_{2} = 3$ , and $c_{2} = 4$ .

Again using orthogonality, $2 (g g_{2} + f f_{2}) = c + c_{2}$ .

So $- 4 g + 6 f = c + 4$ . Subtract the two equations, $(4 g - 6 f) - (- 4 g + 6 f) = (c + 9) - (c + 4)$ .

Hence, $8 g - 12 f = 5$ . Now let the center be $(x, y)$ . Then $g = - x$ and $f = - y$ .

Substitute, $8 (- x) - 12 (- y) = 5$ . So $- 8 x + 12 y = 5$ . Therefore, $8 x - 12 y + 5 = 0$ .

Let the fixed point on the $x$ -axis be $(c, 0)$ . Take any tangent to the circle $x^{2} + y^{2} = a^{2}$ .

Its equation may be written as $y = m x \pm a \sqrt{1 + m^{2}} \Rightarrow m x - y \pm a \sqrt{1 + m^{2}} = 0$ .

Let $(h, k)$ be the foot of the perpendicular from $(c, 0)$ to this tangent. Since $(h, k)$ lies on the tangent,

$m h - k \pm a \sqrt{1 + m^{2}} = 0$ .

Also the line joining $(c, 0)$ to $(h, k)$ is perpendicular to the tangent.

Hence, its slope is $- \frac{1}{m}$ . Therefore, $\frac{k}{h - c} = - \frac{1}{m}$ , so $m = - \frac{h - c}{k}$ .

Substitute in the tangent equation, $- \frac{h (h - c)}{k} - k \pm a \sqrt{1 + \frac{(h - c)^{2}}{k^{2}}} = 0$ .

$\Rightarrow - h (h - c) - k^{2} \pm a \sqrt{k^{2} + (h - c)^{2}} = 0$ .

Transpose, $a \sqrt{(h - c)^{2} + k^{2}} = h (h - c) + k^{2}$ .

Now square and simplify. After reduction, ${(h^{2} + k^{2} - c h)}^{2} = a^{2} ((h - c)^{2} + k^{2})$ .

Replacing $(h, k)$ by $(x, y)$ , the locus is ${(x^{2} + y^{2} - c x)}^{2} = a^{2} ((x - c)^{2} + y^{2})$ .

Let the point $P$ on the circle $x^{2} + y^{2} = 2$ be $(x_{1}, y_{1})$ . Then $x_{1}^{2} + y_{1}^{2} = 2$ .

The tangent at $P$ is $x x_{1} + y y_{1} = 2$ . This tangent cuts the $x$ -axis at the point $L$ .

Putting $y = 0, x x_{1} = 2$ , so $L (\frac{2}{x_{1}}, 0)$ .

Similarly, it cuts the $y$ -axis at the point $M$ . Putting $x = 0, y y_{1} = 2$ , so $M (0, \frac{2}{y_{1}})$ .

Let the midpoint of $L M$ be $(h, k)$ . Then $h = \frac{1}{x_{1}}$ , and $k = \frac{1}{y_{1}}$ .

Therefore, $x_{1} = \frac{1}{h}$ and $y_{1} = \frac{1}{k}$ . Substitute in $x_{1}^{2} + y_{1}^{2} = 2$ .

We get $\frac{1}{h^{2}} + \frac{1}{k^{2}} = 2$ . Hence, the locus is $x^{2} + y^{2} = 2 x^{2} y^{2}$ .

Let the triangle have vertices $(0, 0), (h, 0)$ and $(0, k)$ . Its third side joins $(h, 0)$ and $(0, k)$ .

Hence its equation is $\frac{x}{h} + \frac{y}{k} = 1$ . Or, $k x + h y - h k = 0$ .

This line touches the circle $x^{2} + y^{2} - 2 a x - 2 a y + a^{2} = 0$ .

So the center is $(a, a)$ and the radius is $a$ .

Therefore, the perpendicular distance from $(a, a)$ to the line equals $a$ .

Thus, $\frac{| a k + a h - h k |}{\sqrt{h^{2} + k^{2}}} = a$ .

Squaring, $h k (h k - 2 a h - 2 a k + 2 a^{2}) = 0 \Rightarrow h k - 2 a h - 2 a k + 2 a^{2} = 0$ .

Now the triangle is right-angled at the origin, so the circumcenter is the midpoint of the hypotenuse.

Hence, the circumcenter is $(\frac{h}{2}, \frac{k}{2})$ . Let it be $(x, y)$ . Then $h = 2 x$ , and $k = 2 y$ .

Substitute, $(2 x) (2 y) - 2 a (2 x) - 2 a (2 y) + 2 a^{2} = 0$ . So $4 x y - 4 a x - 4 a y + 2 a^{2} = 0$ .

$\Rightarrow 2 (x + y) - a = \frac{2 x y}{a}$ .

Let the moving point on the circle $x^{2} + y^{2} = 4$ be $(x_{1}, y_{1})$ .

Let the midpoint of $A P$ be $(h, k)$ . Since $A = (1, 5)$ and $P = (x_{1}, y_{1})$ , the midpoint is $h = \frac{x_{1} + 1}{2}$ and $k = \frac{y_{1} + 5}{2}$ .

Hence, $x_{1} = 2 h - 1$ and $y_{1} = 2 k - 5$ . Since $P$ lies on the circle, $x_{1}^{2} + y_{1}^{2} = 4$ .

Substitute, $(2 h - 1)^{2} + (2 k - 5)^{2} = 4 \Rightarrow 2 h^{2} + 2 k^{2} - 2 h - 10 k + 11 = 0$ .

Replacing $(h, k)$ by $(x, y)$ , the locus is $2 x^{2} + 2 y^{2} - 2 x - 10 y + 11 = 0$ .

Let $P (x, y)$ be the midpoint of a variable chord through the fixed point $A (a, b)$ of the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Let the other end of the chord be $Q (x_{1}, y_{1})$ . Since $P$ is the midpoint of $A Q, x_{1} = 2 x - a$ and $y_{1} = 2 y - b$ .

Now $Q$ lies on the circle. Therefore, $(2 x - a)^{2} + (2 y - b)^{2} + 2 g (2 x - a) + 2 f (2 y - b) + c = 0$ .

$\Rightarrow 4 x^{2} - 4 a x + a^{2} + 4 y^{2} - 4 b y + b^{2} + 4 g x - 2 a g + 4 f y - 2 b f + c = 0$ .

$\Rightarrow 4 (x^{2} + y^{2} + g x + f y) - 4 (a x + b y) + a^{2} + b^{2} - 2 a g - 2 b f + c = 0$ .

$=> x^{2} + y^{2} + g x + f y - a x - b y + \frac{a^{2} + b^{2} - 2 a g - 2 b f + c}{4} = 0$ .

This is a circle. The center is $(\frac{a - g}{2}, \frac{b - f}{2})$ . The center of the given circle is $(- g, - f)$ .

The midpoint of the points $(a, b)$ and $(- g, - f)$ is $(\frac{a - g}{2}, \frac{b - f}{2})$ .

Hence, the locus is a circle whose center is the midpoint of the fixed point $A$ and the center of the given circle.

Also, its radius is half the radius of the given circle.

Therefore, the locus is the circle obtained by reducing the given circle in the ratio $1 : 2$ with respect to the point $A$ .

Let the two fixed points be $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ . Let the moving line be $l x + m y + n = 0$ , where $l^{2} + m^{2} = 1$ .

Then the algebraic perpendicular distances of the points from the line are $l x_{1} + m y_{1} + n$ and $l x_{2} + m y_{2} + n$ .

Given that their algebraic sum is constant, say $2 k, (l x_{1} + m y_{1} + n) + (l x_{2} + m y_{2} + n) = 2 k$ .

So $l (x_{1} + x_{2}) + m (y_{1} + y_{2}) + 2 n = 2 k$ . Hence, $n = k - \frac{l (x_{1} + x_{2}) + m (y_{1} + y_{2})}{2}$ .

Substitute this in the equation of the line, $l x + m y + k - \frac{l (x_{1} + x_{2}) + m (y_{1} + y_{2})}{2} = 0$ .

Rearrange, $l (x - \frac{x_{1} + x_{2}}{2}) + m (y - \frac{y_{1} + y_{2}}{2}) + k = 0$ .

Since $l^{2} + m^{2} = 1$ , this represents the tangent form of a circle.

Therefore, the line always touches the fixed circle whose center is $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ and radius is $| k |$ .

Hence, the required fixed circle is ${(x - \frac{x_{1} + x_{2}}{2})}^{2} + {(y - \frac{y_{1} + y_{2}}{2})}^{2} = k^{2}$ .

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Chapter 11. Answers of Circles