10.305. Find the area of the circle circumscribed around a right-angled triangle, the lengths of the legs of which are the roots of the equation $a x^{2}+b x+c=0$.
## Solution.

The radius of the circle circumscribed around a right triangle is equal to half the hypotenuse. Let \( u \) and \( v \) be the lengths of the legs, and \( w \) be the length of the hypotenuse; then \( S = \frac{\pi w^2}{4} \). Since \( u \) and \( v \) are the roots of the equation \( ax^2 + bx + c = 0 \), we have \( uv = \frac{c}{a} \) and \( u + v = -\frac{b}{a} \). Considering that \( w^2 = u^2 + v^2 \), we have \( w^2 = (u + v)^2 - 2uv = \frac{b^2}{a^2} - 2 \frac{c}{a} = \frac{b^2 - 2ac}{a^2} \) and finally find \( S = \frac{\pi (b^2 - 2ac)}{4a^2} \). Answer: \( \frac{\pi (b^2 - 2ac)}{4a^2} \)

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0783.jpg?height=708&width=501&top_left_y=78&top_left_x=210)

Fig. 10.95

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0783.jpg?height=527&width=374&top_left_y=225&top_left_x=774)

Fig. 10.96