Inversion applied on circles I'm studying for my exam and one of the questions I am stuck on is:

Show that under inversion in the unit circle a circle with centre C and radius $S$ inverts into a circle with centre $\frac{C}{C\overline{C} - s^2}$ and radius $\frac{s}{C\overline{C} - s^2}$.

And I know:
$$Z\overline{Z}-Z\overline{C}-\overline{Z}C+C\overline{C}-s^2=0$$
Can someone please explain how to do this question?
 A: Inversion is
$$z\mapsto \frac1{\bar z}$$
Your equation for the circle is basically the squared form of
$$ \lvert z - C\rvert = s $$
Now combine them:
\begin{align*}
\left\lvert\frac1{\bar z}-C\right\rvert&=s \\
\left(\frac1{\bar z}-C\right)\left(\frac1{z}-\bar C\right)&=s^2 \\
\frac1{z\bar z}-\frac{C}{z}-\frac{\bar C}{\bar z}+C\bar C-s^2&=0 \\
1-C\bar z -\bar Cz+z\bar z\left(C\bar C-s^2\right) &= 0 \\
z\bar z - \frac{C}{C\bar C-s^2}\bar z - \overline{\left(\frac{C}{C\bar C-s^2}\right)}z + \frac{1}{C\bar C-s^2} &= 0
\\
\left(z-\frac{C}{C\bar C-s^2}\right) 
\left(\bar z-\frac{\bar C}{C\bar C-s^2}\right)
-\frac{C\bar C}{\left(C\bar C-s^2\right)^2}
+\frac{C\bar C-s^2}{\left(C\bar C-s^2\right)^2}
&= 0
\\
\left(z-\frac{C}{C\bar C-s^2}\right) 
\left(\bar z-\frac{\bar C}{C\bar C-s^2}\right)
-\left(\frac{s}{C\bar C-s^2}\right)^2
&= 0
\\
\left(z-\frac{C}{C\bar C-s^2}\right) 
\left(\bar z-\frac{\bar C}{C\bar C-s^2}\right)
&= \left(\frac{s}{C\bar C-s^2}\right)^2
\\
\left\lvert z-\frac{C}{C\bar C-s^2}\right\rvert
&= \frac{s}{C\bar C-s^2}
\end{align*}
From this you can read
\begin{align*}
C' &= \frac{C}{C\bar C-s^2} &
s' &= \frac{s}{C\bar C-s^2}
\end{align*}
as expected.
