We say two circles $C_1, C_2$ are equidistant if $d(p,C_2)=K$ for any $p \in C_1$ and $d(q,C_1)=K$ for any $q \in C_2$, where $K$ is some constant.
I was told that if I use the Upper-Half plane model (for computation) and later shift the center to a point at infinity, I will be able to prove this simply.
I was able to show that if the center of $C_1, C_2$ (they have the same center) is a normal point (not infinity), then by the definition of the distance between a curve and a point, They are equidistant with the distance $|r_1-r_2|$ where $r_1,r_2$ are radii of circles. (However, this really didn't require the Upper-Half Plane model).
Now, I need to move this center to the infinity point, but I am stuck here. $C_1, C_2$ will become horocycles sharing the same center, and they will still be equidistant because they were just shifted simultaneously.
However, I was unable to get the distant constant $K$ because $r_1 \to \infty, r_2 \to \infty$. Well, I don't have to show the value of $K$, but I need to show that it is constant to claim they are equidistant.
How would the Upper-Half Plane model help show they are equidistant?
Best Answer
I'm assuming that you know that in the Poincaré disk model, a horocycle is represented by a circle which touches the rim of the disk in a single point, the “center” of the horocycle.
You should also know that the Poincaré half plane model relates to the disk model via a Möbius transformation. And Möbius transformations map circles and lines to circles and lines. In this sense, lines can be seen as a special kind of circles, namely those containing the point at infinity. So a horocycle passing though the point at infinity would be a straight line in the half plane model. That's the reason why this specific setup is particularly easy to deal with. The fact that the circle in the disk model is tangent to the model boundary means it contains no other ideal points. Which in the half plane means the line I described earlier does not cross the real axis in any finite point. So it must be a horizontal line.
Now you have simplified your setup to consider two horocycles modelled by horizontal lines in the half plane model. You can either apply formulas to compute distance at any point, or you can reason that a horizontal shift in the model corresponds to a limit rotation which is an isometry. Either way it should be intuitive that the distance is the same all along the horocycle. And since the point at infinity is a special point of the model but not of the underlying hyperbolic geometry, the observation is without loss of generality. It applies to any other pair of concentric horocycles, too.