10,11

In triangle $ABC$, it is known that $AC=12, AB=BC=3\sqrt{10}$. Two spheres touch the plane of triangle $ABC$ at points $A$ and $C$ and are located on opposite sides of this plane. The distance between the centers of these spheres is 15. The center of a third sphere is at point $B$, and this sphere externally touches the two given spheres. Find the radius of the third sphere.
Let $O_1$ and $O_2$ be the centers of the spheres that touch the plane of triangle $ABC$ at points $A$ and $C$ respectively, with radii $x$ and $y$. Let $z$ be the radius of the third sphere. Through the lines $O_1A$ and $O_2C$, which are perpendicular to the plane $ABC$, we draw a plane. In this plane, we drop a perpendicular $O_2P$ from point $O_2$ to the line $O_1A$. In the right triangle $O_1O_2P$, it is known that

$$
O_1O_2 = 15, \quad O_1P = O_1A + AP = O_1A + O_2C = x + y, \quad O_2P = AC = 12.
$$

By the Pythagorean theorem,

$$
(x + y)^2 + 144 = 225
$$

from which we get $x + y = 9$. From the right triangles $ABO_1$ and $CBO_2$, we find that

$$
(x + z)^2 = x^2 + 90, \quad (y + z)^2 = y^2 + 90
$$

from which we get

$$
z^2 + 2xz = 90, \quad z^2 + 2yz = 90
$$

Adding these two equations term by term, we get

$$
2z^2 + 2z(x + y) = 180
$$

and since $x + y = 9$, we have

$$
z^2 + 9z - 90 = 0
$$

from which we find that $z = 6$.

## Answer

6.00