 Solution

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.