Showing that two circles in w-plane tangent each other if there is a Möbius transformation that takes them to two parallel lines
I want to show that two parallel lines in z-plane yield two tanging circles in w-plane. I start with two lines, $L_1=i+x$ and $L_2=3i+x$. Then I want to use the formula:
\begin{equation} \frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} \end{equation}
and having the points on the lines:
$P_1=i, i+1, i+2 $, $P_2=3i,3i+1,3i+2$, I have the right hand side of that equation set. But the left hand side is not so easy to see which points I have to choose. I propose the following for line 1, $L_1=i+x:$
$z_1= i, w_1=i; z_2=i+1, w_2=1; z_3=i+2, w_3=\infty$
I insert in
\begin{equation} \frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} \end{equation}
For the case of line 1:
\begin{equation} \frac{(w-i)(1-\infty)}{(w-\infty)(1-i)}=\frac{(z-i)(i+1-i-2)}{(z-i-2)(i+1-i)} \end{equation}
gives
\begin{equation} \frac{(w-i)}{(1-i)}=\frac{(z-i)(i+1-i-2)}{(z-i-2)(i+1-i)} \end{equation}
which gives the Möbius transformation
\begin{equation} w=\frac{(2 - i) - (1 - 2 i) z}{-2 - i) + z} \end{equation} which has the given w-plot:
Then line 2, $L_2=3i+x$:
$z_1= 3i, w_1=3i; z_2=3i+1, w_2=3i+1; z_3=3i+2, w_3=\infty$
I insert in
\begin{equation} \frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} \end{equation} I insert in
$z_1= 3i, w_1=3i; z_2=3i+1, w_2=3i+1; z_3=3i+2, w_3=\infty$
\begin{equation} \frac{(w-3i)(w_2-\infty)}{(w-\infty)(3i+1-3i)}=\frac{(z-3i)(3i+1-3i-2)}{(z-3i-2)(3i+1-3i)} \end{equation}
which gives
\begin{equation} \frac{(w-3i)}{(3i+1-3i)}=\frac{(z-3i)(3i+1-3i-2)}{(z-3i-2)(3i+1-3i)} \end{equation}
which results in
\begin{equation} w=\frac{(9 - 3 i) - (1 - 3 i) z)}{(-2 - 3 i) + z)} \end{equation}
Which has the following w-plot:
Together, they give:
But I am not so sure the tangent each other. What did go wrong in the choice of w-data?
UPDATE:
By Martins proposition, we take two circles in z-plane, which tangent each other at z=1, $|z|=1 and |z-2|=1$.
We have the new data:
C1: $z_1=0$, $z_2=i$, $z_3=1$; $w_1=0, w_2=i, w_3=\infty$
gives the form of w:
\begin{equation} w=\frac{-zi-z}{(z-1)(i)} \end{equation}
which has the given plot
and the second circle, C2:
C2: $z_1=2, z_2=i+2, z_3=1; w_1=2, w_2=i+2, w_3=\infty$
which gives the form:
\begin{equation} w=\frac{(z-2)(i+1)}{(z-1)}+2 \end{equation}
which has the given plot:
together they give the plot:
If $C_1, C_2$ are two circles in the plane which are tangent to each other then they intersect exactly at one point $w_0 \in \Bbb C$. Then $T(w) = 1/(w-w_0)$ is a Möbius transformation which maps the two circles to two (extended) lines. These lines are parallel because they intersect only at $T(w_0) = \infty$.
Conversely, if $C_1, C_2$ are two circles in the plane and $T$ is a Möbius transformation which maps them to parallel lines then $C_1$ and $C_2$ intersect only at the single point $T^{-1}(\infty)$, which means that they are tangent to each other.