Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 12G / iii / Solution

Rotate the disc so that $P=e^{i\alpha }$ and $Q=e^{-i\alpha }$, with $0<\alpha \le \pi /2$ representing half the smaller angle. For $\alpha <\pi /2$, the hyperbolic line joining them is the arc of a circle orthogonal to the unit circle. Its Euclidean centre is real, say $c$, and orthogonality gives squared radius $c^2-1$. Substituting $P$ into $|P-c{|}^2=c^2-1$ gives $c=\sec \alpha$, with radius $\tan \alpha$.

The arc's point nearest zero is on the positive real axis, atDistance from zero in the Poincare disc model increases with Euclidean modulus, by the radial-length calculation. ThereforeFor $\alpha =\pi /2$, the hyperbolic line is a diameter through zero and its distance is zero, agreeing with the continuous limit of the formula. Choosing the smaller endpoint angle avoids a spurious negative value caused by an inappropriate oriented-angle convention.