= Solution
For distinct points $a,b$ on a <geodesic>, the given disc <isometry> $z\mapsto e^{i\theta}(z-a)/(1-\bar az)$ sends $a$ to zero. Choose $\theta$ so that the image of $b$ is the positive number
$$
x=\left|\frac{b-a}{1-\bar ab}\right|\in(0,1).
$$
An 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 is
$$
d_D(0,x)=\int_0^x\frac{2\,dt}{1-t^2}=2\operatorname{artanh}x.
$$
The diameter is minimizing. Indeed, any path $w(t)$ from zero to $x$ has $|\dot w|\geq|d|w|/dt|$, so its length is at least $2\operatorname{artanh}x$, with equality on the radial segment. Therefore the distance consistent with the given half-plane metric is
$$
\boxed{d_D(a,b)=2\operatorname{artanh}\left|\frac{b-a}{1-\bar ab}\right|.}
$$
For the alternative disc metric $d\widetilde s_D^2=|dw|^2/(1-|w|^2)^2$, every length is halved and the same isometries and <geodesics> remain. In that convention,
$$
\boxed{\widetilde d_D(a,b)=\operatorname{artanh}\left|\frac{b-a}{1-\bar ab}\right|.}
$$
\b[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>.
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