## Proof regarding inversions

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

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

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

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

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