## Proof regarding inversions

Let $I$ be an inversion and let $p$ be a circle such that $I(p)$ is also a circle. When do $p, I(p)$ have equal radii?

My attempt: Let $I_{C,y} = I$ and let $r$ be the radius of $p$, which is fixed, and let $y$ be the radius of $I$. There are then two cases I broke it down into:

If the center $C$ of $I$ lies outside circle $p$. Denote the distance between $C$ and the point on $p$ closest to $C$, $D$, as $x$ and let $E$ be the antipodal point so that $DE$ is a radius of $p$ and $CDE$ are collinear. Since $CD\cdot CI(D) = y^{2}$ we get $CI(D)= \frac{y^{2}}{x}$ and similarly we shall get $CI(E)=\frac{y^{2}}{2r+x}$ and since $I(D)I(E) = CI(E)+CI(D)$ we want $\frac{y^{2}}{x}+\frac{y^{2}}{x+2r} = 2r$ and similarly, if $C$ is within circle $p$ we get $\frac{y^{2}}{x}+\frac{y^{2}}{-x+2r} = 2r$

So if the given variables satisfy either equation, depending on where the center of the inversion is, then $p,I(p)$ will have equal radii.

Is this legitimate? Is there an easier way? Any help would be appreciated, thanks.