[ [ proportional in a right triangle] Common tangent to two circles $\quad]$

Circles $C 1$ and $C 2$ touch externally at point $A$. Line $l$ touches circle $C 1$ at point $B$, and circle $C 2$ at point $D$. Through point $A$, two lines are drawn: one passes through point $B$ and intersects circle $C 2$ at point $F$, and the other is tangent to circles $C 1$ and $C 2$ and intersects line $l$ at point $E$. Find the radii of circles $C 1$ and $C 2$, if $A F=3 \sqrt{2}, B E=\sqrt{5}$.
Let $O_1$ and $O_2$ be the centers of circles $C_1$ and $C_2$ respectively, and $R_1$ and $R_2$ be the radii of these circles. Since $BE = EA = ED$ (as segments of tangents drawn from a single point to a circle), triangle $BAD$ is a right triangle, $\angle BAD = 90^\circ$, and $BD = 2BE = 2\sqrt{5}$. In triangle $DAF$, angle $DAF$ is a right angle, so $DF$ is the diameter of circle $C_2$. Segment $DA$ is the altitude of the right triangle $BDF$ drawn from the vertex of the right angle, so $BD^2 = AB \cdot BF$, or $20 = AB(AB + 3\sqrt{2})$. From this equation, we find that $AB = 2\sqrt{2}$. Then,

$$
4R_2^2 = DF^2 = AF \cdot BF = AF(AF + AB) = 3\sqrt{2}(3\sqrt{2} + 2\sqrt{2}) = 30
$$

Therefore, $R_2 = \frac{\sqrt{30}}{2} = \sqrt{\frac{15}{2}}$. From the similarity of triangles $AO_1B$ and $AO_2F$, we find that

$$
R_1 = AO_1 = AO_2 \cdot \frac{AB}{AF} = R_2 \cdot \frac{2\sqrt{2}}{3\sqrt{2}} = \sqrt{\frac{15}{2}} \cdot \frac{2}{3} = \sqrt{\frac{10}{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-09.jpg?height=766&width=1146&top_left_y=383&top_left_x=502)

Answer

$\sqrt{\frac{10}{3}}, \sqrt{\frac{15}{2}}$