As Poincaré Disk Model is symetric by rotations in the origin (i.e., Rotations through origin are isometries), then hyperbolic circles with center in the origin of the Disk are represented by euclidean circles with same center and vice-versa. This imply that, as conformal circles and hyperbolic circles are invariant through Möbius transformations and Lobachevski transformations (Respectively), then hyperbolic circles are represented by conformal circles of the Hyperbolic Plane and vice-versa. As hyperbolic circles are entirely over the line $r_{\infty}$, then hyperbolic circles are represented by conformal circles that are euclidean circles.
The following shows that the euclidean radius and the hyperbolic radius are different.
Let $C$ be an euclidean circle in the Half-Plane, with center $Oe$.
Let $r$ be the perpendicular to $r_{\infty}$ through $Oe$ and call the intersections with $C$ by $A$ and $B$, where $a<b$ (euclidean coordinates in $r$).
Since $C$ is a circle that is entirely over $r_{\infty}$, it is an hyperbolic circle. So $AB$ is a diameter.
The hyperbolic middle point of $A$ and $B$ is a point located at the $\sqrt{ab}$ coordinate of $r$. Then the euclidean coordinate in $r$ of the hyperbolic center $Oh$ is $a + \sqrt{ab}$.
Since $r$ is a hyperbolic line that is also an euclidean semiline, the hyperbolic radius of $C$ is $\log{\frac{b}{\sqrt{ab}}}$.
Here's a proof which takes advantage of symmetry.
The key fact you need is that Möbius transformations of $\mathbb{C} \cup \{\infty\}$ preserve the set $\{\text{circles}\} \cup \{\text{lines}\}$; this is not hard to prove directly.
It follows that group of Möbius transformations of the upper half plane preserves the set of Euclidean circles in the upper half plane (the only lines in the upper half plane are horizontal and transform to themselves or to Euclidean circles tangent to the real line, none of which is a Euclidean circle entirely contained in the upper half plane). Also, hyperbolic circles are preserved under that action, since that action is the same as the group of orientation preserving isometries of the hyperbolic metric on the upper half plane.
There exists a Möbius transformation that transforms the Poincare disc to the upper half plane, this transformation takes Euclidean circles to Euclidean circles (there are no lines in the Poincare disc), and it takes hyperbolic circles to hyperbolic circles (it is an isometry between the hyperbolic metrics on the Poincare disc and the upper half plane).
So, we have reduced the problem to showing that in the Poincare disc with the hyperbolic metric, hyperbolic circles are the same as Euclidean circles.
The group of Möbius transformations of the Poincare disc equals the group of orientation preserving isometries of the hyperbolic metric on the Poincare disc, and this action is transitive on points, hence for each $r>0$ action is transitive on the set of hyperbolic circles of hyperbolic radius $r$. Thus, it suffices to find, for each $r>0$, a single example of a hyperbolic circle of hyperbolic radius $r$ which is simultaneously a Euclidean circle: and there is evidently a Euclidean circle centered on the origin which is a circle of hyperbolic radius $r$.
Best Answer
One way to show this is to go through the Poincaré disk model. In particular:
The hyperbolic circles in the Poincaré disk model centered at the center point of the disk are clearly the same as the Euclidean circles centered at that point.
Since the isometries of the Poincaré disk model are Möbius transformations, and since Möbius transformations map circles to circles, it follows that the hyperbolic circles centered at any point in the Poincaré disk are Euclidean circles (possibly with a different center).
Finally, since there is a Möbius transformation that maps the Poincaré disk model isometrically to the upper half plane model, it follows that the same holds for the upper half plane model.