Example 10. Draw any line L through the origin, intersecting the circles $x^{2}+y^{2}-2 a x=0$ and $x^{2}+y^{2}-3 a x=0$ at points $A$ and $B$. Take a point $P$ on line $L$ such that $|AP|: |BP| = 1: 2$. Find the equation of the locus of point $P$.
Tears 10
Solution Let the polar equations of the two circles be: $\rho=2 a \cos \theta, p=3 a \cos \theta$ (as shown in Figure 10).

The polar coordinates of points on the same ray L are: $\mathrm{A}\left(\rho_{1}, \theta\right)$, B $\left(\rho_{2}, \theta\right), \mathrm{P}(\rho, \theta)$.

By the given condition $|\mathrm{AP}|:|\mathrm{BP}|=1: 2$, thus, $\left|\rho-\rho_{1}\right|:\left|\rho-\rho_{2}\right|=1: 2$.
Therefore, $\rho=2 \rho_{1}-\rho_{2}$ or $\rho=\frac{1}{3}\left(2 \rho_{1}+\rho_{2}\right)$,

which means $\rho=4 a \cos \theta-3 a \cos \theta=a \cos \theta$,
or $\rho=\frac{7}{3} a \cos \theta$.
Converting to Cartesian coordinates, we get: $x^{2}+y^{2}-a x=0$ or $x^{2}+y^{2}-\frac{7}{3} a x=0$.