9. Given the parabola equation $y=a x^{2}+b x+c(a \neq 0)$, construct a circle $\odot C_{1}$ with radius $\frac{1}{|a|}$ that is internally tangent to the parabola. Construct $\odot C_{2}$ that is externally tangent to $\odot C_{1}$ and internally tangent to the parabola. Construct $\odot C_{3}$ that is externally tangent to $\odot C_{2}$ and internally tangent to the parabola, and so on, resulting in a sequence of circles $C_{1}, C_{2}, C_{3}, \cdots$. Find the sum of the areas of the first $n$ circles.
9. Solution: Without loss of generality, let $a>0$. By the symmetry of the circle and the parabola $y=a x^{2}+b x+c$, and since the circle $C_{n}$ is internally tangent to the parabola, we can assume the equation of the circle $C_{n}$ to be: $\left(x+\frac{b}{2 a}\right)^{2}+\left(y-d_{n}\right)^{2}=r_{n}^{2}$,
$\therefore$ The equation in $y$: $y^{2}+\left(\frac{1}{a}-2 d_{n}\right) y+\left(d_{n}^{2}+\frac{b^{2}}{4 a^{2}}-\frac{c}{a}-r_{n}^{2}\right)=0$ has equal roots,
$\therefore \Delta=\left(\frac{1}{a}-2 d_{n}\right)^{2}-4\left(d_{n}^{2}+\frac{b^{2}}{4 a^{2}}-\frac{c}{a}-r_{n}^{2}\right)=0, \therefore r_{n}^{2}=\frac{b^{2}-1}{4 a^{2}}+\frac{1}{a}\left(d_{n}-c\right)$.
Similarly, we get: $r_{n+1}^{2}=\frac{b^{2}-1}{4 a^{2}}+\frac{1}{a}\left(d_{n+1}-c\right), \quad \therefore r_{n+1}^{2}-r_{n}^{2}=\frac{1}{a}\left(d_{n+1}-d_{n}\right)$.
$\because$ The circles $C_{n}$ and $C_{n+1}$ are externally tangent, $\therefore r_{n+1}+r_{n}=\left|d_{n+1}-d_{n}\right|$.
If $d_{n+1}2)$, there is only one circle $C_{1}$ which does not meet the problem's requirements.
$\therefore d_{n+1}>d_{n}, \therefore r_{n+1}+r_{n}=d_{n+1}-d_{n}, \therefore r_{n+1}-r_{n}=\frac{1}{a}$.
$\therefore\left\{r_{n}\right\}$ is an arithmetic sequence with the first term $r_{1}=\frac{1}{a}$ and common difference $\frac{1}{a}$, $\therefore r_{n}=\frac{n}{a}$.
$\therefore$ The area of the circle $C_{n}$ is: $\frac{\pi}{a^{2}} n^{2}, \therefore$ The sum of the areas of the first $n$ circles is: $\frac{\pi}{a^{2}}\left(1^{2}+2^{2}+\cdots+n^{2}\right)=\frac{\pi}{6 a^{2}} n(n+1)(2 n+1)$.