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.