Let the two known (shared) vertices be $A = (x_A , 0)$ and $B = (x_B , 0)$, and the two unknown vertices be $C = (x_C , y_C)$ and $D = (x_D , y_D)$.
Let the known variables be $x_A$ and $x_B$, and the distances
$$\begin{aligned}
L_{AB} &= x_B - x_A \gt 0 \\
L_{AC} &= \left\lVert\overline{A C}\right\rVert \gt 0 \\
L_{BC} &= \left\lVert\overline{B C}\right\rVert \gt 0 \\
L_{AD} &= \left\lVert\overline{A D}\right\rVert \gt 0 \\
L_{BD} &= \left\lVert\overline{B D}\right\rVert \gt 0 \\
\end{aligned}$$
Then, the system of equations that determines the location of $C$ and $D$ is
$$\left\lbrace \begin{aligned}
L_{AC}^2 &= (x_C - x_A)^2 + y_C^2 \\
L_{BC}^2 &= (x_C - x_B)^2 + y_C^2 \\
L_{AD}^2 &= (x_D - x_A)^2 + y_D^2 \\
L_{BD}^2 &= (x_D - x_B)^2 + y_D^2 \\
\end{aligned} \right .$$
This has four equations and four unknowns, and can be treated as two completely separate systems of equations, each with two unknowns ($x_C$ and $y_C$, and $x_D$ and $y_D$, respectively). The solution is
$$\left\lbrace \begin{aligned}
\displaystyle x_C &= \frac{ L_{AC}^2 - L_{BC}^2 + x_B^2 - x_A^2 }{2 ( x_B - x_A )} \\
\displaystyle y_C &= \pm \sqrt{ L_{AC}^2 - (x_C - x_A)^2 } = \pm \sqrt{ L_{BC}^2 - (x_C - x_B)^2 } \\
\displaystyle x_D &= \frac{ L_{AD}^2 - L_{BC}^2 + x_B^2 - x_A^2 }{2 ( x_B - x_A )} \\
\displaystyle y_D &= \pm \sqrt{ L_{AD}^2 - (x_D - x_A)^2 } = \pm \sqrt{ L_{BD}^2 - (x_D - x_B)^2 } \\
\end{aligned} \right.$$
If the two triangles extend to the same side, then we can choose the positive signs above (since $y_C \gt 0$ and $y_D \gt 0$). Both right sides for the two $y$ coordinates yield the same answer.
After solving $(x_C , y_C)$ and $(x_D , y_D)$ as above, their distance is obviously
$$L_{CD} = \sqrt{ (x_D - x_C)^2 + (y_D - y_C)^2 }$$
Because the distance is necessarily nonnegative, you do not actually need the distance $L_{CD}$ itself; you can just compare the distance squared, $L_{CD}^2$, to the radius squared, $r_C^2$, because nonnegative values compare the same way (less, equal, greater) as their squares do. Expanding the above after squaring yields
$$\begin{aligned}
L_{CD}^2 &= \left( \frac{ L_{AC}^2 + L_{BD}^2 - L_{AD}^2 - L_{BC}^2 }{ 2 (x_B - x_A) } \right)^2 + \left( \sqrt{ L_{AC}^2 - (x_C - x_A)^2 } - \sqrt{ L_{AD}^2 - (x_D - x_A)^2 } \right)^2 \\
~ &= \left( \frac{ L_{AC}^2 + L_{BD}^2 - L_{AD}^2 - L_{BC}^2 }{ 2 (x_B - x_A) } \right)^2 + \left( \sqrt{ L_{BC}^2 - (x_C - x_B)^2 } - \sqrt{ L_{BD}^2 - (x_D - x_B)^2 } \right)^2 \\
\end{aligned}$$
Both yield the same solution, to within numerical precision used.
Note that if you had just placed $A = (0, 0)$, ie. $x_A = 0$, you'd have gotten somewhat simpler solutions (and the math would have been easier, too).
Best Answer
$ HB = \sqrt{ 36-25 }= \sqrt{11} $
$ AH = 9 - \sqrt{11} $
$ AC^2 = AH^2 + HC^2 ; AC = \sqrt{117 -18 \sqrt{11} } $
The area of the triangle is $12×AB×CH$ so you can calculate this easily as suggested by Mark Bennet.
$ R$ = product of sides/ (4 Area)