Reflections of circles through a circle are circles To make things easier, we will try to reflect some general circle through the unit circle. We can use the inverse of the Cayley transform to map our analytic arc in the $z$-plane to the real line in the $\zeta$-plane, apply reflection there, then map it all forward using the Cayley transform. The resulting transformation is $1/\overline{z}$. 
This is where I fumble. I can't figure out how to map a generic circle. Am I to take four points $\lbrace u,v,w,x \rbrace \subset \lbrace z \ |z\in \mathbb{C}, r\in \mathbb{R}, |z-a|=r \rbrace$ and prove that $\lbrace 1/\overline{u},1/\overline{v},1/\overline{w},1/\overline{x} \rbrace$ are concyclic? Thank you all for your time.
 A: The implicit equation of a circle of center $c$ and radius $r$ is
$$r^2=(z-c)\overline{(z-c)}=z\overline z-c\overline z-\overline cz+c\overline c,$$
or
$$z\overline z-c\overline z-\overline cz+d=0$$ where $\color{blue}{d=c\overline c-r^2}.$
Turning $z$ into $1/\overline z$,
$$\frac1{z\overline z}-\frac cz-\frac{\overline c}{\overline z}+d=0.$$
Multiplying by $\dfrac{z\overline z}d$ and rearranging,
$$\frac1d-\frac cd\overline z-\frac{\overline c}dz+z\overline z=0,$$
$$z\overline z-\frac cd\overline z-\frac{\overline c}dz+\frac{c\overline c}{d^2}=\frac{c\overline c}{d^2}-\frac1d.$$
This is a circle of center $\color{blue}{\dfrac cd}$ and radius $\color{blue}{\sqrt{\dfrac{c\overline c}{d^2}-\dfrac1d}}$.
A: I also came up with something. Maybe it works. We know that four points are concyclic if they satisfy $$\frac{(u-v)(w-x)}{(w-v)(u-x)}\in\mathbb{R}$$
Then if we replace $\lbrace u,v,w,x \rbrace$ with $\lbrace 1/\overline{u},1/\overline{v},1/\overline{w},1/\overline{x} \rbrace$ then we get $$\frac{(1/\overline{u}-1/\overline{v})(1/\overline{w}-1/\overline{x})}{(1/\overline{w}-1/\overline{v})(1/\overline{u}-1/\overline{x})}=\overline{\frac{(1/u-1/v)(1/w-1/x)}{(1/w-1/v)(1/u-1/x)}}$$
$$=\overline{\frac{\frac{(v-u)}{vu}\frac{(x-w)}{xw}}{\frac{(v-w)}{vw}\frac{(x-u)}{xu}}} = \overline{\frac{(v-u)(x-w)}{(v-w)(x-u)}}$$
which is justthe same expression that we started with up to subtraction rearrangement, which is only a sign change. Thus this last expression is also real. 
