= Solution
Let $G$ be the <finite group> and choose a point $p$. Its orbit $E=\{gp:g\in G\}$ is a nonempty finite set. Every group element permutes $E$ and maps a closed metric disc onto a disc of the same radius. By the permitted uniqueness of the smallest enclosing disc, it fixes that disc and therefore its centre $o$. Thus $G$ lies in the orientation-preserving stabilizer of one point.
In the <Euclidean plane>, translate $o$ to zero. Every <isometry> fixing zero is an orthogonal <linear map>: distance preservation gives preservation of inner products by the <polarization identity>, and its values on an <orthonormal basis> then determine the <linear map>. Orientation preservation gives <determinant> one, so it is a <rotation>. Consequently $G$ embeds in the circle group of <rotation> angles.
For the <hyperbolic plane>, use this precise standard fact about its <isometries>: after placing a point at the origin of the <Poincare disc model>, every orientation-preserving <isometry> fixing it has the form $z\mapsto e^{i\theta}z$. Equivalently, its point stabilizer is the <rotation> group, and its action on directions at the point is faithful. This is a general hyperbolic-plane fact allowed by the question. It again identifies $G$ with a finite <subgroup> of the circle group.
Finally, a nontrivial finite <rotation> group has a least positive angle $\theta$ in $[0,2\pi)$. Divide any other angle $\phi$ by $\theta$: the remainder $\phi-k\theta$ is another group angle in $[0,\theta)$, so it must be zero. Similarly the remainder of $2\pi$ is zero, and $\theta=2\pi/m$ for an integer $m$. The <rotation> through $\theta$ generates the group. The trivial group is cyclic as well. Therefore \b[both finite orientation-preserving plane <isometry> groups are cyclic]. This proves the <finite orientation-preserving plane isometry groups are cyclic> result using the same enclosing-disc argument in both geometries.
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