Answer: $\frac{20}{47}$ cm and $\frac{45}{47}$ cm.

Solution. Let the radii of the circles be $4x$ and $9x$ respectively, and they touch the legs of the triangle as shown in Figure 2. Express through $x$ the distance $y$ between the points of tangency of the circles with the hypotenuse of the triangle (see Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_06_de2a057edda2939578fcg-2.jpg?height=362&width=1476&top_left_y=293&top_left_x=313)

By the Pythagorean theorem, $(9x + 4x)^2 = (9x - 4x)^2 + y^2$, from which $y = 12x$. Using the double-angle tangent formula, find the tangents of half the angles A and C: $\tan \left(\frac{A}{2}\right) = \frac{1}{3}, \tan \left(\frac{C}{2}\right) = \frac{1}{2}$. Express the length of the hypotenuse AC through $x$: $AC = 42x$. Therefore, $x = \frac{5}{42}$. However, in this case, the right circle will extend beyond the triangle, since the projection of point M on side BC is $\frac{5 \cdot 27}{42} = \frac{135}{42} > 3$ cm from C. This arrangement of the circles does not work. Considering the second option, where the circles are swapped (see Figure 3), gives $x = \frac{5}{47}$ cm. Since $\frac{5 \cdot 36}{47} = \frac{180}{47} < 4$ and $\frac{5 \cdot 12}{47} = \frac{60}{47} < 3$, the circles in this case do not extend beyond the triangle.
11.4. Inside a right-angled triangle with sides 3, 4, and 5 cm, there are two circles, the ratio of whose radii is 9 to 4. The circles touch each other externally, both touch the hypotenuse, one touches one leg, and the other touches the other leg. Find the radii of the circles.