Geodesic in the Poincare half-plane model

The geodesics in the upper half-plane are vertical lines and semicircles centred on the boundary axis. Indeed, the length integrand $\sqrt{1+y^{\prime 2}}/y$ has the Beltrami identitywhich integrates to $(x-c{)}^2+y^2=R^2$. Swapping the coordinate axes gives circles centred on the vertical boundary for the metric $(d{x}^2+d{y}^2)/x^2$.

In the upper half-plane $y>0$, the hyperbolic length element is $ds/y$. The function $\Phi =\mathrm{arsinh}(x/y)$ has gradient $(1,-x/y)/\sqrt{x^2+y^2}$ and norm $1/y$. The Cauchy-Schwarz inequality gives $\int ds/y\ge \Phi (B)-\Phi (A)$ for any oriented path from $A$ to $B$. Equality holds along a semicircle $x^2+y^2=R^2$ traversed with increasing $x$, because its tangent is parallel to that gradient. This is a calibration proof of global minimality, beyond the stationary-path conclusion of the Beltrami identity. Translating $x$ gives the analogous proof for semicircles with other boundary centers.

Two disjoint hyperbolic lines with four distinct ideal endpoints have a unique common perpendicular, and the segment of that perpendicular between them realizes their minimum distance. An upper-half-plane isometry can make the two lines concentric semicircles; their common perpendicular then becomes the imaginary axis.

Three pairwise ultraparallel hyperbolic lines may have disjoint common perpendiculars or coincident ones. In the upper-half-plane model, three semicircles of radius $1/2$ centred at $-5/2,0,5/2$ have mutually disjoint common perpendiculars. Three concentric semicircles instead have the same vertical common perpendicular. Thus pairwise ultraparallelism alone does not determine the incidence of their perpendiculars.

The hyperbolic metric in the half-plane model is conformal, so hyperbolic and Euclidean angles agree. The line meeting the imaginary axis at $iy$, $y>0$, at signed angle $\alpha \in (0,\pi )$ is represented by the Euclidean circle with centre $y\cot \alpha$ on the real axis and radius $y\csc \alpha$.