4. In the Cartesian coordinate system, circle $C_{1}$ intersects with circle $C_{2}$ at points $P$ and $Q$, where the coordinates of point $P$ are $(3,2)$, and the product of the radii of the two circles is $\frac{13}{2}$. If the line $y = kx (k>0)$ and the x-axis are both tangent to circles $C_{1}$ and $C_{2}$, then $k=$ $\qquad$ .
4. $2 \sqrt{2}$ Detailed Explanation: According to the problem, let the centers of the circles be $C_{1}\left(m r_{1}, r_{1}\right)$ and $C_{2}\left(m r_{2}, r_{2}\right)(m>0)$. The equations of circles $C_{1}$ and $C_{2}$ are $\left(x-m r_{1}\right)^{2}+\left(y-r_{1}\right)^{2}=r_{1}^{2},\left(x-m r_{2}\right)^{2}+\left(y-r_{2}\right)^{2}=r_{2}^{2}$. Substituting $(3,2)$ into the two equations, we get $\left(3-m r_{1}\right)^{2}+$ $\left(2-r_{1}\right)^{2}=r_{1}^{2},\left(3-m r_{2}\right)^{2}+\left(2-r_{2}\right)^{2}=r_{2}^{2}$, so $r_{1}$ and $r_{2}$ are the roots of the equation $m^{2} r^{2}-(6 m+4) r+13=0$, thus $m=\sqrt{2}$, so the inclination angle of the line $C_{1} C_{2}$ is $\arctan \frac{\sqrt{2}}{2}$, and the inclination angle of the line $y=k x(k>0)$ is $2 \arctan \frac{\sqrt{2}}{2}$.