Two configurations around same radial line lengths in a triangle? OA is a fixed line length $a$. A circle is drawn tangent at O. A transversal through A cuts the circle at B and C. The radial segment are lengths $(b,c)$.

It is known $ AC=\dfrac{a c}{b} $ by virtue of similar triangles $(OBC,AOC).$
Motivation of the question is a calculation using Cosine Rules etc. suggesting different length combinations of AB & BC.
EDIT1:
This in hindsight appears to be an accuracy problem of CAS that was not expected at start.
Please help finding circumradius in terms of $(a,b,c)$ and angles $(\beta, \gamma)$ between them.
 A: Nope, the circumradius $R$ is unique (if $a,b,c$ allow such a circle to exist).
Choose a coordinate system where $O$ is origin and $A,B,C$ are located at $(a,0)$, $(b\cos\beta,b\sin\beta)$ and $(c\cos\gamma,c\sin\gamma)$ respectively. We will assume $0 < \beta < \gamma \le \frac{\pi}{2}$, this will force $c > b$.
Perform a circle inversion with respect to the unit circle, $A,B,C$ get mapped to
$$A' = \left(\frac1a,0\right),\quad B' = \left(\frac1b\cos\beta,\frac1b\sin\beta\right),\quad C'= \left(\frac1c\cos\gamma,\frac1c\sin\gamma\right)$$
Let $d = 2R$. Since $B,C$ lies on a circle tangent to $OA$ at  $O$. $B', C'$ lie on the line $y = \frac1d$. This implies
$$B' = \left(\sqrt{\frac1{b^2}-\frac1{d^2}}, \frac1d\right)\quad\text{ and }\quad C' = \left(\sqrt{\frac1{c^2}-\frac1{d^2}}, \frac1d\right)$$
Since $A,B,C$ lie on a line, $A',B',C'$ lie on a circle passing through $O$.
Notice $OA' \parallel B'C'$, $OA'$ and $B'C'$ is sharing a common perpendicular bisector. Furthermore, $A'B'$ is mirror image of $OC'$ with respect to this perpendicular bisector. As a consequence, the $x$-coordinates of midpoints of $B'C'$ equals to that of $OA'$. This means
$$\sqrt{\frac1{b^2}-\frac1{d^2}} + \sqrt{\frac1{c^2}-\frac1{d^2}} = \frac1a\tag{*1}$$
As a function of $d$, RHS$(*1)$ isn't real unless $d \ge c = \max(b,c)$.
For $d \in [c,\infty)$, RHS$(*1)$ is strictly increasing and taking values from $\left[\sqrt{\frac1{b^2}-\frac1{c^2}}, \frac1b+\frac1c\right)$.
This means the given geometric configuration is feasible only when
$$\sqrt{\frac1{b^2}-\frac1{c^2}} \le \frac1a < \frac1b + \frac1c$$ and when $a,b,c$ allow such a configuration, the corresponding $d$ and hence $R$ is unique.
With help of a CAS, one find when $R$ exists, it will equal to
$$R = \frac{1}{a\sqrt{\left(\frac1{a^2} + \frac1{b^2} + \frac1{c^2}\right)^2 - 2\left(\frac1{a^4} + \frac1{b^4} + \frac1{c^4}\right)}}$$
