<s>[INST] What is the radius and center of the image of $|z|=1$ under $ f(z) = \frac{3z+2}{4z+3}$? 
I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation:
$$ f(z) = \frac{3z+2}{4z+3} $$
In particular, I'd like to compute the new center and radius.

The Möbius transformation can be turned into inversion as well:

*

*$C_1= 4|z|^2+3\overline{z}-3z-2 $

*$C_2 =|z|^2 - 1$
Or we could turn the second circle into a fractional linear transforation $g(z) = - \frac{1}{z}.$
Then I could multiply the two transformations:
$$
\left[ \begin{array}{cc} 3 & 2 \\ 4 & 3  \end{array} \right]
\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0  \end{array} \right]
=
\left[ \begin{array}{cc} 2 & 3 \\ 3 & 4  \end{array} \right]
 $$
and this could turn back into a circle:

*

*$ C_1C_2 = 3|z|^2 + 4 \overline{z} + 3z + 2 $
I found this technique in a somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation.  I definitely like the idea that Möbius transformations and circles can be identified.
 [/INST] Here's another solution I was able to find.  Notice the matrix factorization:
$$
\left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right]
= 
\left[ \begin{array}{cc} \frac{1}{5} & 0\\ 0 & 5 \end{array} \right] \times 
\left[ \begin{array}{cc} 1 & 18 \\ 0 & 1 \end{array} \right] \times 
\left[ \begin{array}{cr}  \frac{3}{5}& -\frac{4}{5}\\  \frac{4}{5}&  \frac{3}{5} \end{array} \right] = A \times B \times C
 $$
The geometry behind this is that we have a Möbius transformation that factors into three parts:
$$ \text{Möbius} =  rotation \times translation \times dilation $$
Now we have that $|z|=1$ is a circle centered at the origin passing through the points $z = \pm 1$ and $ z = i$.  In fact, all these transformation will map to circles symmetric about the real axis.  Here are the endpoints after the respective transformations:
$$ (-1,1) \stackrel{C}{\to} (7, - \frac{1}{7}) \stackrel{B}{\to} (25,\frac{125}{7})\stackrel{A}{\to} (1, \frac{5}{7}) $$
This corresponds to a circle centered at $z = \frac{6}{7}$ with radius $\frac{1}{7}$.

One possibility for computing this image circle is to notice the circle $|z|=1$ is a geodesic curve in the upper-half plane (with metric $ds^2 = \frac{dx^2 +dy^2}{y^2}$) and passing through the point $(z, \vec{u}) = (i, (1,0)) \in T_1(\mathbb{H}) $.  
A Möbius transformation on $\mathbb{H}$ can be "lifted" to a Möbius transformation on $T_1(\mathbb{H})$ like this:
$$
\left[ z \mapsto \frac{az+b}{cz+d} \right] \to \left[
(z, \vec{u}) \mapsto \left( \frac{az+b}{cz+d} , \frac{\vec{u}}{(cz+d)^2} \right) \right] $$
Let's see what happens when I try the previous example here:
$$ \big(i, (1,0)\big) \mapsto \left( \frac{3i+2}{4i+3}, \frac{(1,0)}{(4i+3)^2}\right) = \left( \frac{18+i}{25} , \frac{1}{25}(24,-7) \right)  $$
The factor of $\frac{1}{25}$ can be discarded since we only need the unit vector.  This map is an isometry in hyperbolic space.  The vector $\vec{u}$ would be tangent to a semi-circle with radius in the direction $\vec{u}_\perp$ passing through the point $f(z)=(\frac{18}{25}, \frac{1}{25})$.  Therefore the center would be:
$$ (\frac{18}{25}, \frac{1}{25}) + \frac{1}{7 \times 25}(24,-7) = (\frac{1}{7},0) $$
agreeing with the previous answer.
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