 Solution

For distinct points $a,b$ on a geodesic, the given disc isometry $z\mapsto e^{i\theta }(z-a)/(1-\bar{a}z)$ sends $a$ to zero. Choose $\theta$ so that the image of $b$ is the positive numberAn orthogonal-circle geodesic through zero is a diameter, so this normalizes the geodesic as required.

There are two hyperbolic metric normalizations to distinguish. For the metric transferred from the stated half-plane, the radial length isThe diameter is minimizing. Indeed, any path $w(t)$ from zero to $x$ has $|\dot{w}|\ge |d|w|/dt|$, so its length is at least $2\mathrm{artanh}x$, with equality on the radial segment. Therefore the distance consistent with the given half-plane metric isFor the alternative disc metric $d{\tilde{s}}_D^2=|dw{|}^2/(1-|w{|}^2{)}^2$, every length is halved and the same isometries and geodesics remain. In that convention,The printed distance corresponds to the latter convention. It is obtained from one quarter of the given half-plane metric, rather than its direct transfer. The equal-point case gives distance zero in both conventions. This is the normalization of the Poincare disc metric.