The parabola $y=a x^{2}+b x+c(a \neq 0)$ intersects the $x$-axis at points $A$ and $B$. The equation of the circle with $AB$ as its diameter is $\qquad$.
8. $a x^{2}+b x+c+a y^{2}=0$.

From the quadratic formula, the diameter $|A B|=\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{|a|}$.
Also, from the axis of symmetry of the parabola, the center is $\left(-\frac{b}{2 a}, 0\right)$. Therefore, the circle with diameter $A B$ is $\left(x+\frac{b}{2 a}\right)^{2}+y^{2}=$ $\frac{b^{2}-4 a c}{4 a^{2}}$, given $a x^{2}+b x+c+a y^{2}=0$.