10.188. The radius of the circle circumscribed around a right triangle relates to the radius of the inscribed circle as 5:2. Find the area of the triangle if one of its legs is equal to $a$.
Solution.

Let $R$ and $r$ be the radii of the inscribed and circumscribed circles, $BC = a$ (Fig. 10.155). Let $BD = x$; then $BL = x$ (as tangents drawn from the same point), $LA = AK = 2R - x$ (since $\triangle ABC$ is a right triangle, then $AB = 2R$). We have $AC^2 + BC^2 = AB^2$ or $(r + 2R - x)^2 + a^2 = 4R^2$. But $R = 5r / 2, x = a - r$, and the last equation will take the form $(7r - a)^2 + a^2 = 25r^2$ or $12r^2 - 7ar + a^2 = 0$, from which $r_1 = a / 3$, $r_2 = a / 4$. These roots correspond to the values $(AC)_1 = 4a / 3$, $(AC)_2 = 3a / 4$. As a result, we get two solutions:

$$
S_1 = \frac{1}{2} a \cdot \frac{4a}{3} = \frac{2a^2}{3}, \quad S_2 = \frac{1}{2} a \cdot \frac{3a}{4} = \frac{3a^2}{8}
$$

Answer: $\frac{2a^2}{3}$ or $\frac{3a^2}{8}$.