= Solution
The curvature-minus-one <Poincaré disk model> has <Riemannian metric>
$$
\boxed{ds^2=\frac{4|dw|^2}{(1-|w|^2)^2}},\qquad |w|<1.
$$
The <geodesics> through zero are Euclidean diameters. Indeed, for a path written in polar coordinates, its length is at least $\int2|\dot r|/(1-r^2)\,dt$, and hence at least $2\operatorname{artanh}r$ between zero and a point of modulus $r$. The radial segment attains this lower bound. By uniqueness of a <geodesic> with given initial tangent, these radial paths describe all <geodesics> through the origin. The radial <hyperbolic distance> is
$$
\rho=\int_0^r\frac{2\,ds}{1-s^2}=\log\frac{1+r}{1-r},
\qquad \boxed{r=\tanh(\rho/2)}.
$$
Thus a <hyperbolic circle in the Poincare disc> is the indicated Euclidean circle.
An <isometry> from the <Poincaré half-plane model> to the <Poincaré disk model> is the <Cayley transform between the half-plane and disk>,
$$
w=\frac{z-i}{z+i},\qquad z=i\frac{1+w}{1-w}.
$$
Writing $z=x+iy$, one has $1-|w|^2=4y/|z+i|^2$ and $|dw|^2=4|dz|^2/|z+i|^4$. Substitution into the disk <Riemannian metric> gives $ds^2=|dz|^2/y^2$, the half-plane <metric>, and $i$ maps to zero.
Put $r=\tanh(\rho/2)$. The inverse image of $|w|=r$ obeys
$$
x^2+(y-1)^2=r^2\bigl(x^2+(y+1)^2\bigr).
$$
Since $(1+r^2)/(1-r^2)=\cosh\rho$, completing the square yields
$$
\boxed{x^2+(y-\cosh\rho)^2=\sinh^2\rho}.
$$
This proves that the <hyperbolic circle in the upper half-plane> has Euclidean centre $i\cosh\rho$ and radius $\sinh\rho$.
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