Find a center and radius of circle that is an image of Mobius Transformation of real axis I need to find a center and radius of a circle that is an image of real axis under homography 
$$ h(z)= \frac{z-z_1}{z-z_2} $$
I found out that homography preserves symetric points, therefore because 
$$ \infty \quad and \quad 0\ $$ are symmetric with respect to generalised circle (real axis) then their images are also symmetric with respect to image of real axis
$$ h(\infty)=1 \quad and \quad h(0)=\frac{z_1}{z_2} $$ 
Any hints or solutions would be appreciated.
 A: The inverse funtion to $f$ is equal $$g(z) =h^{-1} (z) =\frac{z_2 -z_1}{z-1}$$ thus $$f(\{z: \mbox{Im}z =0\} ) =g^{-1} (\{z: \mbox{Im}z =0\})$$
A: That's not quite right. $0$ is symmetric to itself wrt the real line, and ditto for $\infty$. Take the point $z_2$ instead. First consider what happens when $z_2 \in \mathbb R$. If $z_2 \notin \mathbb R$, then you can exploit the symmetry to find the point which is mapped to the center of the circle. The distance between the center and the image of $\infty$ is the radius.
A: Applying Inversive Geometry
$\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}}$
The center of the inversion of $h(z)=\frac{z-z_1}{z-z_2}$ is at $z_2$ and the line, $L$, containing $z_2$ and $\Re(z_2)$ is perpendicular to the real axis. Thus $h(L)$ is a line which is perpendicular to $h(\mathbb{R})$ at both intersections, $h(\infty)$ and $h(\Re(z_2))$. This means that those points are antipodal.
Since
$$
h(\infty)=1\qquad h(\Re(z_2))=\frac{i\Re(z_2)-iz_1}{\Im(z_2)}
$$
we get the radius, $r$ and the center, $k$, of $h(\mathbb{R})$ to be
$$
r=\frac{|z_1-z_2|}{2\Im(z_2)}\qquad k=\frac{i\,\overline{z_2}-iz_1}{2\Im(z_2)}
$$
