Making my comment more explicit ...
Writing $|\cdot|$ for Euclidean distance, and $|\cdot|^\star$ for hyperbolic distance, we have a relatively simple relation for distances from the origin:
$$|OX|^\star = \log\frac{1 + |OX|}{1 - |OX|} = 2 \operatorname{atanh}|OX| \qquad\qquad |OX| = \tanh\frac{\;|OX|^\star}{2} \tag{$\star$}$$
Let the diameter of the target circle meet $\overleftrightarrow{OA}$ at $P$ and $Q$, and define $a := |OA|$, $p := |OP|$, $q := |OQ|$, with $a^\star$, $p^\star$, $q^\star$ their hyperbolic counterparts. Let $r^\star$ be the target circle's hyperbolic radius.
We may assume $p \geq a$ (one of the diameter's endpoints must be on the "far side" of center $A$), so that
$$p^\star = a^\star + r^\star \quad\to\quad p = \tanh\frac{a^\star + r^\star}{2} = \frac{(1+a)\exp r^\star - (1 - a)}{(1+a)\exp r^\star + ( 1 - a)} \tag{1}$$
For the other endpoint, $Q$, an ambiguity arises based on whether the origin lies outside or inside the circle, but we have
$$q^\star = \pm ( a^\star - r^\star ) \quad\to\quad q = \pm \tanh\frac{a^\star - r^\star}{2} = \pm \frac{(1+a) - ( 1 - a )\exp r^\star}{(1+a) + (1-a)\exp r^\star} \tag{2}$$
where "$\pm$" is "$-$" for $O$ inside the circle, and "$+$" otherwise. (If you like, you can absorb the sign into the distances $q$ and $q^\star$, so that they are negative when $\overrightarrow{OA}$ and $\overrightarrow{OQ}$ point in opposite directions, and positive otherwise.)
With the endpoints of the target circle's diameter known, determining the Euclidean center and Euclidean radius is straightforward. $\square$
Note. If $R$ is such that $|OR|^\star = r^\star$, and if we define $r := |OR|$, then $(1)$ and $(2)$ become:
$$p = \frac{a + r}{1 + a r} \qquad\qquad q = \pm \frac{a - r}{1 - a r} \tag{3}$$
I'm assuming that you know that in the Poincaré disk model, a horocycle is represented by a circle which touches the rim of the disk in a single point, the “center” of the horocycle.
You should also know that the Poincaré half plane model relates to the disk model via a Möbius transformation. And Möbius transformations map circles and lines to circles and lines. In this sense, lines can be seen as a special kind of circles, namely those containing the point at infinity. So a horocycle passing though the point at infinity would be a straight line in the half plane model. That's the reason why this specific setup is particularly easy to deal with. The fact that the circle in the disk model is tangent to the model boundary means it contains no other ideal points. Which in the half plane means the line I described earlier does not cross the real axis in any finite point. So it must be a horizontal line.
Now you have simplified your setup to consider two horocycles modelled by horizontal lines in the half plane model. You can either apply formulas to compute distance at any point, or you can reason that a horizontal shift in the model corresponds to a limit rotation which is an isometry. Either way it should be intuitive that the distance is the same all along the horocycle. And since the point at infinity is a special point of the model but not of the underlying hyperbolic geometry, the observation is without loss of generality. It applies to any other pair of concentric horocycles, too.
Best Answer
If you are using the Poincare unit disk model, and you have the unique common perpendicular to the two h-lines, then along that line will be the center of a circle with inversion in the circle exchanging the two h-lines. The circle itself is an h-line and will be fixed, of course, by the inversion. This circle is the locus of points you are looking for because inversion in an h-line is an h-isometry. This situation is analogous to two parallel lines in the Euclidean plane where the locus of points equidistant from both is another parallel line with reflection in this line exchanging the two original parallel lines.