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In the Poincare disc model, a hyperbolic circle in the Poincare disc consists of points at one fixed hyperbolic distance from a center. At center zero it is the Euclidean circle $|z|=\tanh (R/2)$, since radial hyperbolic distance is $2\mathrm{artanh}|z|$. A Poincare disc automorphism sending a general center to zero transports this description. Conversely every Euclidean circle entirely inside the disc is such a circle: the product of inversion in that circle and unit-circle inversion has reciprocal fixed points, one inside; sending the inner fixed point to zero makes the circle concentric with the unit circle.

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 $\sinh \rho$.

For curvature $-1$, the Poincare disc model has metric and area elementThe given minimizing-diameter property makes the distance from zero to Euclidean radius $R$ equal toRadial symmetry therefore identifies the whole hyperbolic circle in the Poincare disc. Integrating its enclosed hyperbolic area gives

For a geodesic hyperbolic triangle with angles $\alpha ,\beta ,\gamma$, the Gauss-Bonnet theorem states $A=\pi -(\alpha +\beta +\gamma )$, in particular $A\le \pi$, with strict inequality for ordinary finite vertices. Let $d$ be the distance of the interior point $P$ to the nearest side. The open disc of radius $d$ about $P$ is contained in the triangle: a path from $P$ to an exterior point must first cross the boundary at distance at least $d$. A hyperbolic isometry carries its center to zero, so its area is the area already calculated. ConsequentlyThis uses hyperbolic area bounds an inscribed disc and establishes the requested bound without assuming that $P$ is an incenter. A sharper universal triangle bound is possible, but is not needed here.

Inversion in a circle of center $c$ and radius $R$ is $K(z)=c+R^2/(\bar{z}-\bar{c})$ on the extended plane. It sends infinity to $c$, so it interchanges zero and infinity exactly when $c=0$.

For the unit-circle inversion $J$, both $J$ and $K$ are involutions. If $KJ{z}_0=z_0$, applying $K$ gives $J{z}_0=K{z}_0$. Writing $w=J{z}_0$, we have $Jw=z_0$ and $Kw=z_0$, and therefore $KJw=K{z}_0=w$. This proves both assertions about the fixed points.

We now show that one fixed point is inside the disk and use it as the hyperbolic center. Rotate coordinates to make the Euclidean center $c=a\ge 0$. The hypothesis is $a+R<1$. If $a=0$ the conclusion follows immediately from the radial hyperbolic distance. If $a>0$, thenfor its fixed points. Since $1+a^2-R^2>2a$, this quadratic has two positive roots with product one, one $p\in (0,1)$ and the other $1/p>1$.

The disk automorphism $h(z)=(z-p)/(1-pz)$ is a Poincare disc automorphism. It sends the two fixed points to zero and infinity and commutes with $J$ on the extended plane. Consequently $K^{\prime }=hK{h}^{-1}=(hT{h}^{-1})J$ interchanges zero and infinity. Conjugation takes circle inversion to inversion in the image circle, so the first part says that $h(\Gamma )$ is centered at zero. Its Euclidean radius is some $s<1$.

In the Poincaré disk metric, integrating the radial length element givesSince $h$ preserves distance, the original circle is exactlyUndoing the preliminary rotation gives the center for arbitrary complex $c$.

This identifies the hyperbolic circle in the Poincare disc intrinsically.