Find a conformal mapping from the unit disc $D$ to itself, such that a given circle $C \subset D$ is mapped to a circle centered at zero

Question

Question in title. Given a circle $C$ centered at $z_0$ with radius $r>0$ that is contained in the open unit disc $D$, can you construct a bijective Mobius transformation from $D$ to itself such that the image of $C$ is a circle centered at zero? I need this to solve a boundary value problem.

Answer 1

By rotation it suffices to consider the case that the center of $C$ lies on the positive real axis, that is, $C:|z-|z_0||=r$.
Let $\alpha =|z_0|+r$ and $\beta =|z_0|-r$. We consider a Mobius transformation $$\varphi (z)=\frac{z-a}{1-az}\quad (0<a<1).$$ The function $\varphi $ satisfies $\varphi (\bar{z})=\overline{\varphi (z)}.$

We determine $a$ so that $\varphi (\alpha )=-\varphi (\beta ).$ Then by the symmetry of $C$ with respect to the real axis and the property of $\varphi $ stated above, $\varphi$ maps $C$ to a circle centered at $0$. We have \begin{align} (\alpha +\beta )a^2-2(\alpha \beta +1)a+\alpha +\beta =0\tag{1} \end{align} from $\varphi (\alpha )=-\varphi (\beta )$. The equation $(1)$ has two positive real roots, one is less than $1$ and the other greater than $1$. Therefore we should take $$a=\frac{\alpha \beta +1-\sqrt{(1-\alpha ^2)(1-\beta ^2)}}{\alpha +\beta }.$$ If we take this $a$, $\varphi$ maps $C$ to a circle centered at $0$ with radius $\varphi (\alpha )=\frac{\alpha -a}{1-a\alpha }$. Since $\varphi $ maps $a$ to $0,$ $a$ is the hyperbolic center of $C$.
In original case, $e^{-i\theta }a$ with $\theta =\arg z_0$ is the hyperbolic center of $C=\{z: |z-z_0|=r\}$.