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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.

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.

For $|\lambda |<1$, let $F_{\lambda }(\zeta )=(\zeta -\lambda )/(1-\overline{\lambda }\zeta )$. Direct calculation givesIts denominator does not vanish in the disc, and this identity shows that it maps the disc into itself. Its inverse $(w+\lambda )/(1+\overline{\lambda }w)$ has the same property, so $F_{\lambda }$ is a Poincare disc automorphism. Multiplication by a complex number of modulus one is another such automorphism.

Conversely let $T$ be a Möbius transformation mapping the disc onto itself, and let $\lambda =T^{-1}(0)$. Then $T\circ F_{\lambda }^{-1}$ is a disc automorphism fixing zero, say $S(\zeta )=a\zeta /(c\zeta +d)$ with $d\ne 0$. Since it maps the boundary circle to itself, $|a{|}^2=|c\zeta +d{|}^2=|c{|}^2+|d{|}^2+2\mathrm{Re}(c\overline{d}\zeta )$ for every $|\zeta |=1$. This forces $c\overline{d}=0$, hence $c=0$ and $|a/d|=1$. Thus $S$ is a rotation $\eta \zeta$, and $T=\eta F_{\lambda }$. Absorb the denominator's minus sign into $\omega =-\eta$ to obtain precisely the paper's convention:Here mapping the disc to itself means bijectively onto itself; there are nonautomorphic Möbius transformations whose image is a proper subdisc.

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.

A suitable Poincare disc automorphism is the Möbius transformationIt sends $a$ to zero and has inverse $(w+a)/(1+\overline{a}w)$. To verify the isometry directly, note thatThese identities show both that $f$ maps the unit disc onto itself and that $2|df|/(1-|f{|}^2)=2|dz|/(1-|z{|}^2)$, preserving the Riemannian metric.