Horocycle transformation in the Poincare half plane model I was puzzeling with how to find an easy formula to calculate the length of a horocycle in the Poincare half plane model 
Then I had the brainwave that I can just use a transformation and then find the formula from that
My brainwave in detail:


*

*Given the points $P_1 (x_1, y_1) $ and $P_2 (x_2, y_2) , P_1 \not= P_2, (y_{1,2}  \gt 0) $

*Find the transformation that moves:
$P_1 \to (0, 1) $ and $P_2 \to (z, 1)  $ with  $ z \gt 0 $

*The length of the horocycle is then simply $z$
I can move $P_1 \to (0, 1) $. 
But how to get in the same transformation move $P_2 \to (z, 1) $?
For people more into the complex plane (i am not so familiar with it) 
how to move
$ p_1 \to i $ and $p_2 \to (z + i ) $ with z being real in a single mobius transformation
 A: Here's how to do it in a couple of steps, which can easily be converted into an analytic formula. I'll assume $P_1 \ne P_2$.


*

*In the Euclidean plane, let $C$ be a circle contained in the closed upper half plane, passing through $P_1$ and $P_2$, and tangent to the real line. There will always be exactly two such circles. Let $Q$ be the point of tangency in the real line. 

*Apply a Mobius transformation taking $Q \to \infty$. The image of the circle $C$ is now a horizontal line $y=c$ (all Mobius transformations have real coordinates, as befit isometries of the hyperbolic metric on the upper half plane).

*Compose with the Mobius transformation $w \mapsto \frac{1}{c} w$ (I'll use $w=x+iy$ as my complex variable). The image of $C$ is now the line $y=1$, and the the images of $P_1$ and $P_2$ both lie on this line, with $P_1 \mapsto (x',1)$

*Compose with the Mobius transformation $w \mapsto w - (x'+i)$.


Thus, by composing three Mobius transformations, you will get the Mobius transformation of your desire.
