We take the two perpendicular lines as axes of the coordinates. Let $(x,y)$ be any point satisfying the given condition. According to condition $x+y=a$.

This is the relation which connects the coordinates of any point on the locus is the equation to the locus.

In the next chapter we will study that it is an equation of a straight line.

Let $(x,y)$ be any position of the moving point. By given condition from the question we have

$(x-a{)}^2+y^2+(x+a{)}^2+y^2=2c^2\Rightarrow x^2+y^2=c^2-a^2$

This is the relation between coordinates of each and every point that satisfies the given condition, and is the equation to required locus.

The equation tells us that the square of distance of the point $(x,y)$ from the origin is constant and equal to $c^2-a^2$, and therefore, the locus of the point is a circle whose center is origin and radius is $\sqrt{c^2-a^2}$.

Let the point be $P(x,y)$. According to question

$\sqrt{(x+1{)}^2+y^2}=3\sqrt{x^2+(y-2{)}^2}\Rightarrow (x+1{)}^2+y^2=9(x^2+(y-2{)}^2)$

$\Rightarrow 8(x^2+y^2)-2x-36y+35=0$.

Let the point be $P(x,y)$. Given that $P{A}^2-P{B}^2=2k^2$

$\Rightarrow (x-a{)}^2+y^2-(x+a{)}^2-y^2=2k^2\Rightarrow -4ax=2k^2\Rightarrow 2ax+k^2=0$.

Given that $PA=nPB\Rightarrow P{A}^2=n^{2}P{B}^2\Rightarrow (x-a{)}^2+y^2=n^2[(x+a{)}^2+y^2]$

$\Rightarrow (n^2-1)(x^2+y^2+a^2)+2ax(n^2+1)=0$.

Given that $PA+PB=c\Rightarrow \sqrt{(x-a{)}^2+y^2}+\sqrt{(x+a{)}^2+y^2}=c$

Squaring, $\Rightarrow (x-a{)}^2+y^2+(x+a{)}^2+y^2+2\sqrt{(x-a{)}^2+y^2}\sqrt{(x+a{)}^2+y^2}=c^2$

Squaring again, $\Rightarrow 4x^2(c^2-4a^2)+4c^2{y}^2=c^4-4a^2$.

Given that $P{B}^2+P{C}^2=2P{A}^2\Rightarrow (x+a{)}^2+y^2+(x-c{)}^2+y^2=2[(x-a{)}^2+y^2]$

Simplifying gives us $(6a-2c)x=a^2-c^2$.

Let $P(x,y)$ be the point in question. Its distance from $y$-axis would be $x$. Thus, according to question

$(x-1{)}^2+(y-2{)}^2=x^2\Rightarrow y^2-4y-2x+5=0$.

Let $P(x,y)$ be the point in question, then according to the question

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

Let $P(x,y)$ be the point in question, then according to the question

$(x-2{)}^2+(y-3{)}^2=(x-4{)}^2+(y-5{)}^2\Rightarrow 4x+4y=28\Rightarrow x+y=7$.

Let $P(x,y)$ be the point in question, then according to the question

$(x-a-b{)}^2+(y-a+b{)}^2=(x-a+b{)}^2+(y-a-b{)}^2\Rightarrow x=y$.

Distance from $x$-axis is $y$ and distance from $y$-axis is $x$, then according to the question $y=3x$.

$(x-a{)}^2+y^2=4^2.x^2\Rightarrow 15x^2-y^2+2ax=a^2$.

According to question, $x+y=3$.

According to question, $x^2+(y-2{)}^2=4\Rightarrow x^2+y^2-4y=0$.

According to question, $(x-3{)}^2+y^2=3^2[x^2+(y-2{)}^2]\Rightarrow 8x^2+8y^2+6x-36y+27=0$.

$y=\frac{1}{2}\sqrt{x^2+y^2}\Rightarrow x^2=3y^2$.

Let the fixed straight line be parallel to the $x$-axis, and let the fixed point be the origin $O(0,0)$.

Since the fixed point is at a perpendicular distance $a$ from the line, and the line is parallel to the $x$-axis, its equation must be $y=a$.

We consider moving point as $P(x,y)$.

Its distance from origin is $\sqrt{x^2+y^2}$. Its distance from fixed line is $|y-a|$.

Thus, $x^2+y^2=(y-a{)}^2=>x^2+2ay=a^2$.

From given condition $(x^2+y^2)=\frac{1}{4}(y-a{)}^2\Rightarrow 4x^2+3y^2+2ay=a^2$ and $\frac{1}{4}(x^2+y^2)=(y-a{)}^2\Rightarrow x^2-3y^2+8ay=4a^2$.

Let $P=(x,y)$, then $x=a\cos \theta ,y=b\sin \theta$

$\cos \theta =\frac{x}{a},\sin \theta =\frac{y}{b}\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Let $P(x,y)$ be the required point, then according to the question

$\sqrt{x^2+(y-2{)}^2}+\sqrt{x^2+(y+2{)}^2}=6\Rightarrow 9x^2+5y^2=45$.

Let $AB=2a$, where $A$ is $(-a,0)$ and $B$ is $(a,0)$. Let $P(x,y)$ be the moving point. Then according to the question

$A{P}^2+P{B}^2=A{B}^2\Rightarrow (x+a{)}^2+y^2+(x-a{)}^2+y^2=4a^2\Rightarrow x^2+y^2=a^2$.

Tip: The obtained equation is a circle(we will see this in later chapters) which is evident because from diameters the angle on any point on the perimeter is a right angle.

Figure 8.1. 

The diagram is given below:

Let $A=(h,0),B=(0,k)$ and $P=(x,y)$. Given $AP=a,PB=b$.

Now $\frac{AP}{PB}=\frac{a}{b}$, hence, $P$ divides $AB$ in the ration $a:b$.

Thus, $x=\frac{a.0+b.h}{a+b}=\frac{bh}{a+b},y=\frac{a.k+b.0}{a+b}=\frac{ak}{a+b}$

Also, $A{B}^2=O{A}^2+O{B}^2\Rightarrow (a+b{)}^2=h^2+k^2=\frac{(a+b{)}^2{x}^2}{b^2}+\frac{(a+b{)}^{2}2y^2}{a^2}$

$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Figure 8.2. 

Let $A=(x_1,y_1)$. Let the moving point be $P(\alpha ,\beta )$ then according to the question $A(x_1,y_1)$ lies on the curve $y^2=8x$.

$\Rightarrow y_1^2=8x_1$. Also, since $P$ is the mid-point of $OA$,

$\alpha =\frac{x_1}{2}\Rightarrow x_1=2\alpha$ and $\beta =\frac{y_1}{2}\Rightarrow y_1=2\beta$

Thus, $4{\beta }^2=8.2.\alpha \Rightarrow {\beta }^2=4\alpha$. Thus, locus of point $P$ is $y^2=4x$.

Let $f(x,y)=x^2+y^2-2x+1.f(2,-5)=26\ne 0$.

Thus, the point $(2,-5)$ does not lie on the given curve.

Since the given equation represent the identical curves the ratio of coefficients of terms must be equal. Thus,

$\frac{m_1{m}_2}{a}=-\frac{m_1+m_2}{2h}=\frac{1}{b}\Rightarrow m_{1}m-2=\frac{a}{b},m_1+m_2=\frac{2h}{b}$.

Let $(x,y)=(a{t}^2,2at)$. Then $y^2=4ax$.

Let $(x,y)=(t+\frac{1}{t},t-\frac{1}{t})$, then $x^2-y^2=4$.

Let $(x,y)=(\cos \theta +\sin \theta ,\cos \theta -\sin \theta )$, then $x^2+y^2=2$.

Centroid will be given by $(\frac{\cos \theta +\sin \theta +1}{3},\frac{\sin \theta +\cos \theta +2}{3})$. Let it be $(x,y)$.

Let $\cos \theta +\sin \theta =S$, then $x=\frac{S+1}{3}\Rightarrow 3x=S+1$ and $3y=S+2$.

Subtracting we get $3x-3y+1=0$.

Let $P(u\cos \alpha .t,u\sin \alpha .t-k{t}^2)=(x,y)$ be the moving point. Then

$x=u\cos \alpha .t\Rightarrow t=\frac{x}{u\cos \alpha }$ and $y=u\sin \alpha .t-k{t}^2$

Substituting for $t$ in $y$ we have $y=x\tan \alpha -\frac{k{x}^2}{u^2{\cos }^2\alpha }$.

Let $P(x,y)$ be the point whose locus is to be determined, then according to the question

$(x+2{)}^2+(y-3{)}^2=3^2[(x^2+(y-3{)}^2)]\Rightarrow 8x^2-4x-4+8(y-3{)}^2=0$

$\Rightarrow 2x^2+2y^2-x-12y+17=0$.

Let $P(x,y)$ be the point whose locus is to be determined, then according to the question

$4P{A}^2=9P{B}^2\Rightarrow 4[(x+5{)}^2+(y-3{)}^2]=9[(x-2{)}^2+(y-4{)}^2]$

$\Rightarrow 5x^2+5y^2-76x-48y+44=0$.

Let $P(x,y)$ be the point whose locus is to be determined, then according to the question

$PS=\sqrt{(x-4{)}^2+y^2},PM=x$ and $PS=PM\Rightarrow (x-4{)}^2+y^2=x^2$

$\Rightarrow y^2-8x+16=0$.

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 $PA=PB$.

$P{A}^2=(x-x_1{)}^2+(y-y_1{)}^2$ and $P{B}^2=(x-x_2{)}^2+(y-y_2{)}^2$

Since $PA=PB\Rightarrow (x-x_1{)}^2+(y-y_1{)}^2=(x-x_2{)}^2+(y-y_2{)}^2$

$x^2-2x{x}_1+x_1^2+y^2-2y{y}_1+y_1^2=x^2-2x{x}_2+x_2^2+y^2-2y{y}_2+y_2^2$

$-2x{x}_1-2y{y}_1+x_1^2+y_1^2=-2x{x}_2-2y{y}_2+x_2^2+y_2^2$

Rearranging, $2x(x_2-x_1)+2y(y_2-y_1)=x_2^2+y_2^2-x_1^2-y_1^2$

This is a linear equation in $x$ and $y$, hence it represents a straight line.

Now let $M$ be the midpoint of $AB$. Then $M(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

Substituting the coordinates of $M$ into the equation satisfies it, so the line passes through the midpoint of $AB$.

Now we consider triangles $PAM$ and $PBM$ and we see that $PA=PB,PM=PM,AM=BM$ so both the triangles are congruent. We also find that $P{A}^2=P{M}^2+A{M}^2$, thus, the triangles are right angled as well at $M$. Thus, we have proven that the locus bisects $AB$ at right angle.

Distance of $P(x,y)$ from $(a,0)$ is $\sqrt{(x-a{)}^2+y^2}=a+x$

Sqauring gives us, $\Rightarrow (x-a{)}^2+y^2=(x+a{)}^2\Rightarrow y^2=4ax$.

Let $A(1,2),B(-2,3)$ and $P(x,y)$.

The area of triangle $PAB$ is given by $\mathrm{\Delta }=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Substituting $A(1,2),B(-2,3)$ and $P(x,y)$

$\mathrm{\Delta }=\frac{1}{2}|x(2-3)+1(3-y)+(-2)(y-2)|\Rightarrow |-x-3y+7|=18$

$x+3y=-11$ or $x+3y=25$.

Let $A(1,2)$ and let $Q(x,y)$ be a variable point on the curve $x^2+y^2+x+y=0$.

Let $P(h,k)$ be the midpoint of $AQ$. By the midpoint formula,

$h=\frac{1+x}{2},k=\frac{2+y}{2}$

Hence, $x=2h-1,y=2k-2$.

Since $Q$ lies on the curve, $(2h-1{)}^2+(2k-2{)}^2+(2h-1)+(2k-2)=0$

$\Rightarrow 4h^2-4h+1+4k^2-8k+4+2h-1+2k-2=0$

$\Rightarrow 2h^2+2k^2-h-3k+1=0$

So the locus is $2x^2+2y^2-x-3y+1=0$.

If $P(x,y)$ divides $AQ$ internally in the ratio $3:1$, then by section formula,

$x=\frac{3.3\cos \theta +1.2}{4},y=\frac{3.2\sin \theta +1.3}{4}$.

$x=\frac{9\cos \theta +2}{4},y=\frac{6\sin \theta +3}{4}$.

Rearranging, $4x-2=9\cos \theta$, $4y-3=6\sin \theta$.

Thus, $\cos \theta =\frac{4x-2}{9}$, $\sin \theta =\frac{4y-3}{6}$.

Using ${\cos }^2\theta +{\sin }^2\theta =1$,

${\left(\frac{4x-2}{9}\right)}^2+{\left(\frac{4y-3}{6}\right)}^2=1$.

Hence the locus of $P$ is $\frac{(4x-2{)}^2}{81}+\frac{(4y-3{)}^2}{36}=1$.

Let a variable line through $A(6,-8)$ cut the $x$-axis at $Q(a,0)$.

Let $P(x,y)$ be the midpoint of $AQ$.

By midpoint formula, $x=\frac{6+a}{2},y=\frac{-8+0}{2}$.

So, $x=\frac{6+a}{2},y=-4$.

From the first equation, $a=2x-6$.

Since $a$ is arbitrary (any real number), $x$ can take any real value, while $y$ is constant.

Hence the locus of $P$ is $y=-4$.