I have the following problem:
Reflect the line $y=x$ in the circle $|z-2|=1$.
I know that the map $$z\mapsto\frac{R^2}{\overline{z}-\overline{a}}+a$$ take a point to a symmetric point respect to the circle $|z-a|=R$.
So, in this case $\infty\mapsto2$ and $(1+i)\mapsto\frac{3+i}{2}$.
Both points are in the line $x+y=2$. To find the reflection I think that I have to consider the system of equations given by
$$2=\frac{r^2}{2-\overline{a}}+a,$$ $$\frac{3+i}{2}=\frac{r^2}{\frac{3-i}{2}-\overline{a}}+a.$$
I know that the answer is $$\left|z-\frac{7+i}{4}\right|=\frac{\sqrt{2}}{4},$$ but from the previous equations I can't get there. What I'm missing?
Any help will be appreciated.
By symmetry, if $z$ is a point on the image, then the reflectional image of $z$ is on the line.
$y=x$ can be written as $z=i\bar z$. The image of $z$ is $\displaystyle \frac{1}{\bar z-2}+2$. So we have
\begin{align*} \frac{1}{\bar z-2}+2&=i\left(\frac{1}{ z-2}+2\right)\\ \frac{2\bar z-3}{\bar z-2}&=\frac{i(2z-3)}{z-2}\\ 2z\bar z-3z-4\bar z+6&=i(2z\bar z-4z-3\bar z+6)\\ z\bar z-\frac{3-4i}{2(1-i)}z-\frac{4-3i}{2(1-i)}\bar z+3&=0\\ z\bar z-\frac{7-i}{4}z-\frac{7+i}{4}\bar z+3&=0\\ \left(z-\frac{7+i}{4}\right)\left(\bar z-\frac{7-i}{4}\right)&=\frac{1}{8}\\ \left|z-\frac{7+i}{4}\right|&=\frac{\sqrt{2}}{4} \end{align*}