= Solution
For the <Poincare disc model>, radial distance from zero to Euclidean radius $r$ is
$$
\int_0^r\frac{2\,ds}{1-s^2}=2\operatorname{artanh}r.
$$
It 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. Thus
$$
\boxed{\text{every hyperbolic circle in the disc model is a Euclidean circle}.}
$$
This argument does not say that their Euclidean and hyperbolic centres coincide.
Back to article page
