6. Given that the lengths of the two legs of a right triangle are the two roots of the quadratic equation $a x^{2}+b x+c=0$. Then the area of the circumcircle of the right triangle is $\qquad$ (the result should be expressed using $a, b, c$ and $\pi$).
6. $\frac{\pi\left(b^{2}-2 a c\right)}{4 a^{2}}$.

Let the lengths of the two legs of the right triangle be $x_{1}$ and $x_{2}$. Then
$$
x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a} .
$$

The hypotenuse of the right triangle is
$$
\begin{array}{l}
\sqrt{x_{1}^{2}+x_{2}^{2}}=\sqrt{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}} \\
=\sqrt{\frac{b^{2}-2 a c}{a^{2}}} .
\end{array}
$$

Thus, the radius of the circumcircle of the right triangle is
$$
R=\frac{1}{2} \sqrt{\frac{b^{2}-2 a c}{a^{2}}} \text {. }
$$

Therefore, the area of the circumcircle of the right triangle is
$$
\pi R^{2}=\frac{\pi\left(b^{2}-2 a c\right)}{4 a^{2}} .
$$