= Solution
Use curvature $-1$. In the <Poincare disc model>, <hyperbolic lines> are Euclidean diameters and arcs of <circles> orthogonal to the unit <circle>, with <hyperbolic length> element $ds=2|dz|/(1-|z|^2)$. In the <upper half-plane model>, they are vertical lines and semicircles with centres on the real axis, with $ds=|dz|/\operatorname{Im}z$. Both displayed <hyperbolic metrics> are positive scalar multiples of the Euclidean metric, so their <angles> agree with Euclidean <angles>. The <hyperbolic distance> is the length of the joining <hyperbolic line> segment; explicitly, in the <upper half-plane model>,
$$
\boxed{d(P,Q)=\operatorname{arcosh}\left(1+\frac{|P-Q|^2}{2\operatorname{Im}P\operatorname{Im}Q}\right)}.
$$
In the <Poincare disc model> it is $2\operatorname{artanh}|(P-Q)/(1-\overline P Q)|$.
For distinct $P,Q$, a <isometry> sends their joining <hyperbolic line> to the imaginary axis, so their images are $ip,iq$ with $p,q>0$. For any <continuously differentiable> curve $x(t)+iy(t)$ joining them,
$$
\ell(\gamma)=\int\frac{\sqrt{x'(t)^2+y'(t)^2}}{y(t)}\,dt
\geq\int\frac{|y'(t)|}{y(t)}\,dt
\geq|\log q-\log p|=d(P,Q).
$$
Equality in the first inequality requires $x'=0$ everywhere, because the nonnegative difference is <continuous>. Equality in the second requires $y'$ to have one sign, allowing zero intervals. Thus \b[equality holds precisely for a monotone reparametrisation of the joining hyperbolic segment]. Conversely every such reparametrisation gives equality. For $P=Q$, equality means length zero and the constant curve; monotonicity is understood non-strictly.
Two distinct <hyperbolic lines> are <parallel hyperbolic lines> if they are disjoint in the plane and have exactly one common <ideal endpoint>; they are <ultraparallel hyperbolic lines> if they are disjoint and have no common <ideal endpoint>. To prove the <common perpendicular of ultraparallel hyperbolic lines> theorem, send one line to the imaginary axis. An <ultraparallel hyperbolic lines> second line can, after reflection if necessary, be written as a semicircle with centre $c>0$ and radius $r$ satisfying $c>r$. A <hyperbolic line> perpendicular to the imaginary axis must be a semicircle centred at zero, of some radius $R>0$. The Euclidean condition for its orthogonality to the second <circle> is
$$
\boxed{R^2=c^2-r^2}.
$$
This has exactly one positive solution, proving existence and uniqueness. Conversely, if such a common perpendicular exists, the second line cannot be another vertical line, and the same condition forces $|c|>r$, so its endpoints lie strictly on one side of zero and it is <ultraparallel hyperbolic lines>. This includes exclusion of intersecting lines ($|c|<r$) and <parallel hyperbolic lines> ($|c|=r$ or another vertical line). The statement concerns distinct lines: a line coincident with itself would have many perpendiculars.
A <horocycle> in the <upper half-plane model> is either a Euclidean <circle> tangent to the real axis from above, with the tangent point as its <ideal centre of a horocycle>, or a horizontal line $y=h>0$, whose <ideal centre of a horocycle> is infinity. A <isometry> sending this centre to infinity sends the <horocycle> to a horizontal line. The <hyperbolic lines> meeting that horizontal line orthogonally are exactly the vertical lines; therefore the <hyperbolic lines> orthogonal to a <horocycle> are exactly those with its ideal centre as an endpoint.
If two <horocycles> have distinct <horocycle centres>, the unique <hyperbolic line> with those two <ideal endpoints> meets both orthogonally. If they have the same <horocycle centre>, send it to infinity: both become horizontal lines and every vertical <hyperbolic line> meets both orthogonally. Hence
$$
\boxed{\text{The common orthogonal line is unique exactly when the ideal centres differ.}}
$$
In the other case there are infinitely many; intersection or tangency of the two <horocycles> does not change this classification.
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