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In the Poincaré half-plane model, a hyperbolic circle of radius $R$ centred at $u+iv$ is the Euclidean circle $(x-u{)}^2+(y-v\cosh R{)}^2=v^2{\sinh }^{2}R$. Its Euclidean centre is shifted upward from its hyperbolic centre, and its Euclidean radius is $v\sinh R$. The lower height $v{e}^{-R}$ is positive. The formula follows by mapping the centre to zero in the Poincare disc model and using radial distance $2\mathrm{artanh}\rho$.

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 $a$.

In the upper half-plane model, a hyperbolic line is either a vertical line or a Euclidean semicircle perpendicular to the real axis. For the vertical line $x=a$, set $R_L(z)=2a-\bar{z}$. For the circle $|z-a|=r$ with real $a$, setThe vertical formula preserves $y$ and Euclidean speed. For the circular formula, $\mathrm{Im}{R}_L(z)=r^2\mathrm{Im}z/|z-a{|}^2$, while its local Euclidean length multiplier is $r^2/|z-a{|}^2$. Thus both preserve $|dz|/\mathrm{Im}z$, the hyperbolic metric, and consequently all curve lengths and distances. Each is an involution preserving the half-plane; each fixes its specified line pointwise and moves points off that line. This establishes the required hyperbolic reflection explicitly.

To factor an arbitrary isometry $g$, first recall why a bisector reflection exchanges any distinct points $P,Q$. The hyperbolic distance formula isEquating the distances to $P$ and $Q$ yields a vertical line or a circle with centre on the real axis, hence a hyperbolic line. Substitution in the reflection formulas above shows that its reflection exchanges $P$ and $Q$. Choose a base point $P$. If $gP\ne P$, compose $g$ on the left with their bisector reflection, obtaining $h$ that fixes $P$; otherwise take $h=g$.

Choose $Q\ne P$. The points $hQ,Q$ have the same distance from $P$, so their bisector contains $P$. If they differ, its reflection corrects $hQ$ while retaining $P$. We obtain an isometry $k$ fixing both $P,Q$ after at most two reflections.

An isometry fixing two points is either the identity or reflection in their joining line. To see this directly, a point with prescribed distances from $P,Q$ lies at the intersection of two hyperbolic circles; these are Euclidean circles in this model and have at most two intersections. When there are two, reflection in the line $PQ$ interchanges them. Pick a third point off that line. Either $k$ fixes it, or composing $k$ with that line reflection fixes it. An isometry fixing three noncollinear points is the identity: the first two distances leave only the reflected pair, and the distance to the third point distinguishes that pair. Thus $k$ is the identity or one additional reflection. Reversing the compositions proves every isometry is a product of at most three hyperbolic reflections.

For the Poincare disc model, radial distance from zero to Euclidean radius $r$ isIt is minimal: any curve has length at least the integral of $2|dr|/(1-r^2)$, which in turn is at least the net radial integral. Consequently a hyperbolic circle of radius $a$ centred at zero is $|\zeta |=\tanh (a/2)$, a Euclidean circle.

Given a centre $p$, use a Poincare disc automorphism carrying $p$ to zero. Since it is an isometry, it takes the given hyperbolic circle to the preceding Euclidean circle. A Möbius transformation takes generalized circles to generalized circles. The preimage here is compact and contained in the disc, so it cannot be a straight line. ThusThis argument does not say that their Euclidean and hyperbolic centres coincide.

An ideal hyperbolic triangle has its three vertices at infinity, with sides given by geodesics. In curvature $-1$ its angles are zero and its area is $\pi$, using the hyperbolic triangle area formula $\pi -(\alpha +\beta +\gamma )$.

To calculate the area of a hyperbolic disc, use the Poincare disc model centered at the origin. Hyperbolic radial distance is $\rho ={\int }_0^{R}2dr/(1-r^2)=2\mathrm{artanh}R$, so its Euclidean radius is $R=\tanh (\rho /2)$. The metric area density is $4rdrd\theta /(1-r^2{)}^2$, givingEvery hyperbolic triangle, including ones with ideal vertices, has area at most $\pi$. But $A(2)=2\pi (\cosh 2-1)>\pi$. A hyperbolic triangle is geodesically convex, so if it contains a complete hyperbolic circle, it also contains the enclosed hyperbolic disc, its geodesic convex hull. This would violate the area comparison. Therefore no hyperbolic triangle contains a complete circle of radius two.

In the curvature $-1$ normalization, the hyperbolic triangle area is its angle defect:Join one vertex of a convex geodesic polygon to its nonadjacent vertices. This gives $n-2$ hyperbolic triangles with disjoint interiors. Their angles at every polygon vertex sum to that vertex's interior angle. Adding the hyperbolic triangle areas therefore gives the hyperbolic polygon areaFor a regular hyperbolic polygon with prescribed area, place $n$ equally spaced vertices on a hyperbolic circle of radius $R>0$ and join consecutive vertices by geodesics. Rotations through $2\pi /n$ and reflections in radial lines show that this is a convex regular polygon. The triangle from its centre to a vertex and the midpoint of an adjacent side is right angled, with angles $\pi /n$, $\alpha (R)/2$, and $\pi /2$, and hypotenuse $R$. The angle form of the hyperbolic law of cosines givesConsequently $A(R)=(n-2)\pi -n\alpha (R)$ is continuous and strictly increasing. As $R\downarrow 0$, $\alpha (R)\to \pi -2\pi /n$, so $A(R)\to 0$; as $R\to \infty$, $\alpha (R)\to 0$, so $A(R)\to (n-2)\pi$. The intermediate value theorem proves existence for every permitted $A$. Indeed the required radius isThe argument also proves uniqueness of the radius in this construction; neither endpoint area is attained by a nondegenerate finite-radius polygon.