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The Poincaré half-plane model has Riemannian metric $d{s}^2=(d{x}^2+d{y}^2)/y^2=|dz{|}^2/(\mathrm{Im}z{)}^2$. Every Möbius transformation preserving this half-plane can be represented as $f(z)=(Az+B)/(Cz+D)$ with real coefficients and $\Delta =AD-BC>0$. Direct calculation givesTherefore $|df|/\mathrm{Im}f=|dz|/\mathrm{Im}z$, proving it is an isometry.

The imaginary axis is a hyperbolic geodesic. Its distance from $ib$ to $ic$ is $|{\int }_b^{c}dy/y|=|\log (c/b)|$. This is also the minimum over all paths, since path length is at least $|\int dy/y|$. The two vertical points at distance $r$ from $ib$ are $ib{e}^{-r}$ and $ib{e}^r$. Reflection in the imaginary axis preserves the hyperbolic circle and its centre. Since it is assumed to be a Euclidean circle, its Euclidean centre lies on that axis and its two vertical intersections determine the Euclidean centre and radius:This proves the hyperbolic circle in the upper half-plane formula for the specified centre.

For the horizontal-distance calculation, the geodesic joining $ib$ to $a+ib$ is the semicircle of centre $a/2$ and radius $R=\sqrt{b^2+a^2/4}$. Parameterize it as $z=a/2+R{e}^{i\theta }$. Its metric length is $|d\theta |/\sin \theta$. If the right endpoint angle is ${\theta }_0$, then $\sin {\theta }_0=b/R$, $\cos {\theta }_0=a/(2R)$, and the other endpoint angle is $\pi -{\theta }_0$. HenceThe equality of horizontal hyperbolic and Euclidean distances therefore requiresLet $t=a/2>0$. The derivative of $t/\sinh t$ has the sign of $\sinh t-t\cosh t$. This quantity vanishes at zero and has derivative $-t\sinh t<0$, so the height function is strictly decreasing. Its limits are one at $a\downarrow 0$ and zero at $a\to \infty$. Continuity and the intermediate value theorem prove that each $0<b<1$ has exactly one positive matching separation $a$.