= 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, at
$$
r_* =\sec\alpha-\tan\alpha=\frac{1-\sin\alpha}{\cos\alpha}.
$$
Distance from zero in the <Poincare disc model> increases with Euclidean modulus, by the radial-length calculation. Therefore
$$
\boxed{d(O,PQ)=2\operatorname{artanh}\!\left(\frac{1-\sin\alpha}{\cos\alpha}\right).}
$$
For $\alpha=\pi/2$, 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.
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