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This is exercise 1.11 in the book Fuchsian groups by Svetlana Katok.

Show that every hyperbolic circle in the upper half plane $\mathbb{H}$ is a Euclidean circle (with a different center, of course), and vice versa.

I am aware of the fact that we could work with the ball model as suggested in this answer, but I think that the author's intent was to use the upper half plane model. Has anyone a hint for me showing this?

I tried to consider first a circle around $i$ since in the ball model we consider first circles around $0$, but until now without success.

Edit. We may cheat a bit. In a book on hyperbolic geometry I found the following:

Let $A$ be a circle in the upper half plane $\mathbb{H}$. Suppose the Euclidean centre of $A$ is $a + ib$ and the Euclidean radius of $A$ is $r$. Then the hyperbolic centre is $a + i\sqrt{b^2 - r^2}$ and the hyperbolic radius $R$ satisfies $r = b\mathrm{tanh}(R)$

So we just have to verify this. I mean, if $A$ is above, we have that $A$ is parametrized by $a + ib + re^{it}$. Thus we have to compute $$\rho(a + i\sqrt{b^2 - r^2}, a + ib + re^{it})$$ where $\rho$ is the hyperbolic distance function. I have the formula $$\mathrm{tanh}\left( \frac{1}{2}\rho(z,w)\right) = \frac{|z - w|}{|z - \bar{w}|}$$ However this gets quite complicated...

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  • $\begingroup$ See the answer I have given in (math.stackexchange.com/q/2211803). Don't be frightened by the equations of geodesics. See further with the description of a multi-layer optical model. $\endgroup$ – Jean Marie Sep 23 '17 at 21:45
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    $\begingroup$ Euclidean circles that are tangent or cut the boundary line are not hyperbolic circles (tangent circles are horocycles, cutting circles are lines or hypercycles) $\endgroup$ – Willemien Sep 24 '17 at 12:35
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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 $r$.

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