How to find the hyperbolic center of a circle in the complex plane?

Question

I need this for a boundary value problem. I am asked to map the right half plane with a closed disk (bounded by a circle $C$) cut out of it conformally to an annulus. My plan of action is to map the half plane to the unit disc with a mobius transformation $M$, then compose that with a mobius transformation $T$ that maps the unit disc to itself such that $T(M(C))$ is centered at zero. To do that i need to map the hyperbolic center of $M(C)$ to zero. How do I find the hyperbolic center of $M(C)$?

Answer 1

If $C=\{z:|z-z_0|=r\}$ with $z_0=x_0+iy_0\,(0<r<x_0)$, then the hyperbolic center of $C$ is $a=\sqrt{{x_0}^2-r^2}+iy_0$ and $$ \varphi (z)=\frac{z-a}{z+\bar{a}} $$ maps $C$ to a circle centered at the origin. We can employ the same method as in this.

Answer 2

Assuming you're in the Poincaré disk model, you can simply take the line connecting the Euclidean center of the circle to the center of the Poincaré disk. That line intersects the circle in two points, and the hyperbolic midpoint of these is the hyperbolic center of the circle. Finding the midpoint can be done by parametrizing the line and setting the hyperbolic distances as equal. There will be multiple solutions, but only one which is real and lies inside the unit disk.

Instead of computing the center as described, you could also consider hyperbolic translations along the line through the Euclidean center. Parametrize these, then find one which maps the two points of intersection to a pair of points symmetric around the center of the Poincaré disk. The inverse of that operation will map the origin to the hyperbolic center of the original circle $M(C)$, but you no longer need that if the circle is already centered, right?

NB: this answer on Math Overflow was useful to me when dealing with circles under Möbius transformations. So if you (or some other reader) don't have a ready solution for how to find the center of $M(C)$ itself, that might help.