The hypotenuse $AB$ has length $AB = 5$ with respect to Pythagoras' theorem. Then for angles in the triangle there is $\cos \alpha = \frac{4}{5}$, $\cos \beta = \frac{3}{5}$,
$$\cot \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = 3,$$
$$\cot \frac{\beta}{2} = \sqrt{\frac{1 + \cos \beta}{1 - \cos \beta}} = 2.$$

Since both circles $k_1, k_2$ whole lie in the triangle $ABC$, they are externally tangent—in the opposite case the leg tangent to the smaller circle intersects the greater circle. Let circles $k_1$ and $k_2$ touch the side $AB$ at points $D$ resp. $E$ and let point $F$ be orthogonal projection of the point $S_2$ to the element $S_1D$ (Fig. 1, under assumption is $r_1 > r_2$). Using Pythagoras' theorem for a triangle $FS_2S_1$ we obtain
$$(r_1 + r_2)^2 = (r_1 - r_2)^2 + DE^2,$$
which follows $DE = 2\sqrt{r_1r_2}$.
An equality $AB = AD + DE + EB$ gives
$$c = r_1 \cot \frac{\alpha}{2} + 2\sqrt{r_1r_2} + r_2 \cot \frac{\beta}{2} = 3r_1 + 2\sqrt{r_1r_2} + 2r_2,$$
and since $r_1 = \frac{9}{4}r_2$ we obtain
$$\frac{27}{4}r_2 + 3r_2 + 2r_2 = 5,$$
which follows
$$r_2 = \frac{20}{47}, \quad r_1 = \frac{45}{47}.$$