Question

Show that the locus of a point, which is such that the tangents from it to two given concentric circles are inversely as the radii, is a concentric circle, the square of whose radius is equal to the sum of the squares of the radii of the given circles.

Answer 1

For our convenience, we take the concentric circles to be centred at origin with radius $R,r$ such that $R>r$.

C1: $x^2+y^2=R^2$

C2: $x^2+y^2=r^2$

Let the point which satisfies the condition be $(h,k)$.

length of tangent from point $(a,b)$ to circle $x^2+y^2=k^2$ is:

$l=\sqrt{x^2+y^2-k^2}$

According to the question:

$\frac{R}{r}=\frac{\sqrt{h^2+k^2-r^2}}{\sqrt{h^2+k^2-R^2}}$

$⟹R^2(h^2+k^2-R^2)=r^2(h^2+k^2-r^2)$

$⟹(x^2+y^2+R^2+r^2)=0$

Hence Proved.