17 (12 points) In the Cartesian coordinate system, it is known that circle $C_{1}$ intersects with circle $C_{2}$ at points $P, Q$, and the coordinates of point $P$ are $(3, 2)$. The product of the radii of the two circles is $\frac{13}{2}$. If circles $C_{1}$ and $C_{2}$ are both tangent to the line $l: y=k x$ and the $x$-axis, find the equation of the line $l$.
17 From the problem, we know that $O, C_{1}, C_{2}$ are collinear. Let the radii of circles $C_{1}$ and $C_{2}$ be $r_{1}, r_{2}$, respectively, and the slope of line $C_{1} C_{2}$ be $\tan \alpha \neq 0$. Let $m = \cot \alpha$, then the centers of circles $C_{1}$ and $C_{2}$ are $C_{1}\left(m r_{1}, r_{1}\right), C_{2}\left(m r_{2}, r_{2}\right)$, and the equations of the two circles are
$$
\begin{array}{l}
\left(x-m r_{1}\right)^{2}+\left(y-r_{1}\right)^{2}=r_{1}^{2}, \\
\left(x-m r_{2}\right)^{2}+\left(y-r_{2}\right)^{2}=r_{2}^{2} .
\end{array}
$$

Point $P(3,2)$ is a common point of the two circles, so
$$
\begin{array}{l}
\left(3-m r_{1}\right)^{2}+\left(2-r_{1}\right)^{2}=r_{1}^{2}, \\
\left(3-m r_{2}\right)^{2}+\left(2-r_{2}\right)^{2}=r_{2}^{2} .
\end{array}
$$

From this, we know that $r_{1}, r_{2}$ are the roots of the equation
$$
m^{2} r^{2}-(6 m+4) r+13=0
$$

Thus, $r_{1} r_{2}=\frac{13}{m^{2}}, m=\sqrt{2}$. Therefore, the equation of line $l$ is
$$
y=\tan 2 \alpha \cdot x=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha} x=2 \sqrt{2} x .
$$