Answer: $\frac{4 \sqrt{55}}{11}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_dc247db770ca7d1ba777g-5.jpg?height=502&width=717&top_left_y=1551&top_left_x=664)

Solution. Point $O$ is the midpoint of $A C$. Since the center of the circumscribed circle of triangle $A B C$ lies on its median $B O$, this median coincides with the perpendicular bisector of side $A C$, and triangle $A B C$ is isosceles with base $A C$.

Angle $A D C$ is a right angle as an inscribed angle subtending the diameter of the circle. In triangle $A B C$, two altitudes are drawn - $B O$ and $A D$. Therefore, triangle $D O C$ is similar to triangle $A B C$ with a similarity coefficient $k=\cos \angle A C B=\frac{D C}{A C}=\frac{3}{8}$. By Heron's formula,

$$
p_{D O C}=\frac{2 D O+D C}{2}=\frac{11}{2}, \quad S_{D O C}=\sqrt{\frac{11}{2} \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{5}{2}}=\frac{\sqrt{55}}{2} .
$$

Therefore,

$$
r_{A B C}=\frac{r_{D O C}}{k}=\frac{3 \sqrt{55}}{4}: \frac{11}{2}: \frac{3}{8}=\frac{4 \sqrt{55}}{11}
$$
9. (30 points) On the side $AC$ of triangle $ABC$, a circle is constructed with center $O$, intersecting the side $BC$ of this triangle at point $D$. Find the radius of the inscribed circle of triangle $ABC$, if $OD=4$, $DC=3$, and the center of the circumscribed circle of triangle $ABC$ lies on $BO$.