= Solution
Inversion in a circle of center $c$ and radius $R$ is $K(z)=c+R^2/(\bar z-\bar c)$ on the extended plane. It sends infinity to $c$, so \b[it interchanges zero and infinity exactly when $c=0$].
For the unit-circle inversion $J$, both $J$ and $K$ are involutions. If $KJz_0=z_0$, applying $K$ gives $Jz_0=Kz_0$. Writing $w=Jz_0$, we have $Jw=z_0$ and $Kw=z_0$, and therefore $KJw=Kz_0=w$. This proves both assertions about the <fixed points>.
We now show that one <fixed point> is inside the disk and use it as the hyperbolic center. Rotate coordinates to make the Euclidean center $c=a\geq0$. The hypothesis is $a+R<1$. If $a=0$ the conclusion follows immediately from the radial hyperbolic distance. If $a>0$, then
$$
T(z)=a+\frac{R^2z}{1-az},\qquad
az^2-(1+a^2-R^2)z+a=0
$$
for its <fixed points>. Since $1+a^2-R^2>2a$, this quadratic has two positive roots with product one, one $p\in(0,1)$ and the other $1/p>1$.
The disk automorphism $h(z)=(z-p)/(1-pz)$ is a <Poincare disc automorphism>. It sends the two <fixed points> to zero and infinity and commutes with $J$ on the extended plane. Consequently $K'=hKh^{-1}=(hTh^{-1})J$ interchanges zero and infinity. Conjugation takes circle inversion to inversion in the image circle, so the first part says that $h(\Gamma)$ is centered at zero. Its Euclidean radius is some $s<1$.
In the Poincaré disk metric, integrating the radial length element gives
$$
d_D(0,z)=\int_0^{|z|}\frac{2\,dr}{1-r^2}=2\operatorname{artanh}|z|.
$$
Since $h$ preserves distance, the original circle is exactly
$$
\boxed{\Gamma=\{z\in D:d_D(p,z)=2\operatorname{artanh}s\}.}
$$
Undoing the preliminary rotation gives the center for arbitrary complex $c$.
This identifies the <hyperbolic circle in the Poincare disc> intrinsically.
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