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

By the quadratic formula, the diameter of the circle is
$$
|A B|=\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{|a|} .
$$

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

which is $a x^{2}+b x+c+a y^{2}=0$.