If we write $\;\alpha=a+ib\,,\,\;a,b\in\Bbb R\;$ , then
$$\frac1{\overline\alpha}=\frac1{a-ib}=\frac{a+ib}{a^2+b^2}=\frac\alpha{a^2+b^2}$$
and then we get: $\;\alpha\;$ is on the first circle and $\;\frac1{\overline\alpha}\;$ is on the second one:
$$\begin{cases}(i)\;\,\;\;(a-x_0)^2+(b-y_0)^2=r^2\\{}\\(ii)\;\;\left(\cfrac a{a^2+b^2}-x_0\right)^2+\left(\cfrac b{a^2+b^2}-y_0\right)^2=4r^2\end{cases}$$
and we're also given that $\;2(x_0^2+y_0^2)=r^2+2\;$ , then we need quite a few substitutions and stuff:
$$(i)\;\;a^2+b^2-2(ax_0+by_0)+\overbrace{x_0^2+y_0^2}^{=\frac12r^2+1}=r^2\implies a^2+b^2-2(ax_0+by_0)=\frac12r^2-1$$
But also
$$(ii)\;\;\frac{a^2+b^2}{(a^2+b^2)^2}-2\left(\frac{ax_0+by_0}{a^2+b^2}\right)+\overbrace{x_0^2+y_0^2}^{=\frac12r^2+1}=4r^2\implies \frac{a^2+b^2}{(a^2+b^2)^2}-2\left(\frac{ax_0+by_0}{a^2+b^2}\right)=\frac72r^2-1$$
And from here we get that
$$\begin{cases}(i)\implies-2(ax_0+by_0)=\cfrac12r^2-1-a^2-b^2\\{}\\
(ii)\implies-2(ax_0+by_0)=\left[\cfrac72r^2-1- \cfrac{a^2+b^2}{(a^2+b^2)^2}\right](a^2+b^2)=\cfrac{7r^2-2}2(a^2+b^2)-1\end{cases}$$
and comparing right sides:
$$\left(\frac72r^2-1\right)(a^2+b^2)-1=\frac12r^2-1-a^2-b^2\implies\frac72r^2(a^2+b^2)=\frac12r^2\implies...$$
Finish now the argument.