The image of a circle under $ z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1} $ Let $C_1 = \{ z: |z| = \frac{1}{7}  \}$ be a circle inside the unit circle $C_0 = \{ z: |z|=1\}$.  The fractional linear transformation:
$$ \phi: z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1} $$
is a map from the unit cirle to itself, $\phi: C_0 \to C_0$.  What is the image of $C_1$ ?  Obviously it is a circle, but what is the radius and center of $\phi(C_1)$?

I hope this figure not too confusing.  $\phi$ maps the light blue circle to the dark blue circle.  Similarly it maps light green to dark green.



In regard to 3 points... I know there is formula for 3 points $(x_1, y_1), (x_2, y_2),(x_3, y_3)$ going through a circle using a determinant but I have not used it:

$$ \left|\begin{array}{cccc} x^2 + y^2 &amp; x &amp; y &amp; 1 \\ 
x_1^2 + y_1^2 &amp; x_1 &amp; y_1 &amp; 1 \\
x_2^2 + y_2^2 &amp; x_2 &amp; y_2 &amp; 1 \\
x_3^2 + y_3^2 &amp; x_3 &amp; y_3 &amp; 1 \\ \end{array} \right|=0$$

Which 3 points should I pick anyway?  Even more alternatively, are there ways to solve this using power of a point from geometry?



I suspect @vvnitram's answer is wrong.  Here is a plot of the image circle (computed numerically) and the circle he proposes.  These are not quite the same.
Let $$w=u+iv=\frac{2z-1}{z-2}$$

Rearranging, we get $$z=\frac{2w-1}{w-2}$$

So the image of the circle $|z|=\frac 17$ is given by $$\left|\frac{2w-1}{w-2}\right|=\frac 17$$
$$\Rightarrow 7|2w-1|=|w-2|$$
$$\Rightarrow 49((2u-1)^2+4v^2)=(u-2)^2+v^2$$

This simplifies to $$195u^2+195v^2-192u+45=0$$

This is a circle, centre $(\frac{32}{65},0)$ and radius $\frac {7}{65}$
Note that $\phi(0)=1/2$ and $\phi(1/7)=5/13$.

Because is a moebious transform, send the center into center. Then, $\phi(C_1)=C$ where $C$ is the circunference with center $1/2$ and radious $1/2-5/13=3/26$