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The curvature-minus-one Poincaré disk model has Riemannian metricThe geodesics through zero are Euclidean diameters. Indeed, for a path written in polar coordinates, its length is at least $\int 2|\dot{r}|/(1-r^2)dt$, and hence at least $2\mathrm{artanh}r$ between zero and a point of modulus $r$. The radial segment attains this lower bound. By uniqueness of a geodesic with given initial tangent, these radial paths describe all geodesics through the origin. The radial hyperbolic distance isThus a hyperbolic circle in the Poincare disc is the indicated Euclidean circle.

An isometry from the Poincaré half-plane model to the Poincaré disk model is the Cayley transform between the half-plane and disk,Writing $z=x+iy$, one has $1-|w{|}^2=4y/|z+i{|}^2$ and $|dw{|}^2=4|dz{|}^2/|z+i{|}^4$. Substitution into the disk Riemannian metric gives $d{s}^2=|dz{|}^2/y^2$, the half-plane metric, and $i$ maps to zero.

Put $r=\tanh (\rho /2)$. The inverse image of $|w|=r$ obeysSince $(1+r^2)/(1-r^2)=\cosh \rho$, completing the square yieldsThis proves that the hyperbolic circle in the upper half-plane has Euclidean centre $i\cosh \rho$ and radius $\sinh \rho$.

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 $a$.

First obtain the hyperbolic circle in the upper half-plane. For centre $w=u+iv$, the fractional-linear map $\zeta =(z-w)/(z-\overline{w})$ takes the Poincaré half-plane model isometrically to the Poincare disc model, sending $w$ to zero. Direct differentiation transforms the metric to $4|d\zeta {|}^2/(1-|\zeta {|}^2{)}^2$. A radial segment from zero to modulus $\rho$ has length $2\mathrm{artanh}\rho$. Any path to that modulus has at least this length, by $|d\zeta |\ge |d\rho |$; thus it is the hyperbolic distance.

A circle of hyperbolic radius $R$ therefore satisfiesSquaring and rearranging givesHence it is a circle in Euclidean geometry, with Euclidean centre $u+iv\cosh R$ and radius $v\sinh R$. Its lowest point has height $v{e}^{-R}>0$, so it stays entirely in the upper half-plane. The centres differ whenever $R>0$: for example, the hyperbolic circle centred at $i$ has Euclidean centre $i\cosh R$, rather than $i$.

For hyperbolic side-side-side congruence, use the standard fact that hyperbolic isometries act transitively on ordered point pairs with the same positive separation, and that reflection in a hyperbolic line is an isometry. Map $A,B$ to $A^{\prime },B^{\prime }$. The image of $C$ has prescribed distances $b$ from $A^{\prime }$ and $a$ from $B^{\prime }$, just as $C^{\prime }$ does. There are at most two such points, interchanged by reflection in the line $A^{\prime }{B}^{\prime }$.

For completeness, normalize $A^{\prime },B^{\prime }$ to points on the positive imaginary axis. Their distance circles are Euclidean circles with centres on that axis by the formula above. Subtracting their equations determines one height $y$, and then the abscissa is determined up to its sign. Reflection $z\mapsto -\overline{z}$ interchanges these choices. Therefore either the chosen isometry or its composition with that reflection maps all three labelled vertices correctly. Equal three side lengths imply a hyperbolic isometry between the triangles.

Two sides and an unspecified equal angle do not suffice. Here is an explicit hyperbolic side-side-angle ambiguity. Fix the angle at $A$ to be $\pi /6$, let $\cosh a=3/2$ and $\cosh b=2$, and chooseBoth are positive and distinct. The hyperbolic cosine rule at $A$ becomesSetting $u=e^c$, the condition is $u^2-6u+7=0$, whose two roots are $3\pm \sqrt{2}$. Construct each triangle from sides $b,c_{\pm }$ meeting at angle $\pi /6$; its opposite side is the same $a$ by the cosine rule. The triangles therefore share $a,b,A$ but have different third sides and cannot be isometric. If the common angle is specifically the included angle between the two given sides, the cosine rule determines the third side and side-side-side congruence does apply.