9.1. Two spheres touch the plane of triangle $A B C$ at points $A$ and $B$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 7, and the distance between their centers is 13. The center of a third sphere with radius 5 is at point $C$, and it touches each of the first two spheres externally. Find the radius of the circle circumscribed around triangle $A B C$.
Answer. $\sqrt{30} \approx 5.48$.

Solution. Let $O_{1}$ and $O_{2}$ be the centers, and $r_{1}$ and $r_{2}$ be the radii of the spheres that touch the plane $A B C$ at points $B$ and $A$ respectively, $r_{3}=5$ be the radius of the third sphere. By the problem's condition, $r_{1}+r_{2}=r=7$, $O_{1} O_{2}=d=13$. Let also $A B=c, B C=a, A C=b$. From triangles $O_{1} B C$ and $O_{2} A C$ using the Pythagorean theorem, we get

$$
\left(r_{3}+r_{1}\right)^{2}=r_{1}^{2}+a^{2}, \quad\left(r_{3}+r_{2}\right)^{2}=r_{2}^{2}+b^{2}
$$

from which $a^{2}+b^{2}=2 r_{3}^{2}+2\left(r_{1}+r_{2}\right) r_{3}$. Using the arrangement of points $O_{1}, B, A$ and $O_{2}$, we find

$$
\left(r_{1}+r_{2}\right)^{2}+c^{2}=d^{2}
$$

Therefore,

$$
a^{2}+b^{2}-c^{2}=2 r_{3}^{2}+2\left(r_{1}+r_{2}\right) r_{3}+\left(r_{1}+r_{2}\right)^{2}-d^{2}=2 \cdot 5^{2}+2 \cdot 7 \cdot 5+7^{2}-13^{2}=0
$$

Thus, triangle $A B C$ is a right triangle, and the radius of the circumscribed circle around it is half the hypotenuse, i.e.,

$$
R=\frac{c}{2}=\frac{\sqrt{d^{2}-\left(r_{1}+r_{2}\right)^{2}}}{2}=\frac{\sqrt{d^{2}-r^{2}}}{2}=\sqrt{30} \approx 5.48
$$