Question

Show that the locus of points such that the ratio of its distances from two given points is constant, is a circle. Hence show that the circle cannot pass through the given points.

Answer 1

Let the given points be chosen along x-axis and the distance between them be $2a$ and their mid-point as origin.

Hence their coordinates are

$A(a,0),.B(-a,0).$

Let P be any point $(x,y)$ such that

$\frac{PA}{PB}=k.$

∴$P{A}^2=k^{2}P{B}^2$

or $(x-a{)}^2+y^2=k^2[(x+a{)}^2+y^2]$

or $(x^2+y^2+a^2)(1-k^2)-2ax(1+k^2)=0$

or $x^2+y^2+a^2-2ax$ ⋅ $\frac{1+k^2}{1-k^2}=0$

Which is a circle.

It will pass through (a, 0) if

$2a^2-2a^2$⋅ $\frac{1+k^2}{1-k^2}=0$ or $4a^2{k}^2=0$

Above is not possible as $a\ne 0,k\ne 0$

Therefore the point A does not lie on the circle.

Now putting $(-a,0),$ we get

$2a^2+2a^2\cdot$ ,$\frac{1+k^2}{1-k^2}=0$ or $4a^2=0$

Above is also not possible as $a\ne 0$.

Hence $B$ also does not lie on the circle