13.3.11 ** $A, B$ are two given points on a plane, $C$ moves on a circle with $A$ as the center, and the internal angle bisector of $\triangle ABC$ at angle $A$ intersects side $BC$ at point $P$. Find the locus of point $P$.
Parse: With $A$ as the origin of coordinates, $AB$ as the $x$-axis, establish a rectangular coordinate system. Let the circle centered at $A$ be the unit circle $x^{2}+y^{2}=1$. Let $C\left(x_{0}, y_{0}\right), B(a, 0), P(x$, $y)$. Since $AP$ is the angle bisector of $\angle CAB$, then $\frac{AB}{AC}=\frac{BP}{PC}=$ $\frac{a}{1}=a$, thus by the section formula we get $x=\frac{a+a x_{0}}{1+a}$, $y=\frac{a y_{0}}{1+a}$. Therefore, we solve $\left\{\begin{array}{l}x_{0}=\frac{(1+a) x-a}{a}, \\ y_{0}=\frac{(1+a) y}{a} .\end{array}\right.$
Since $x_{0}^{2}+y_{0}^{2}=1$, we have $\left[\frac{(1+a) x-a}{a}\right]^{2}+\left[\frac{(1+a) y}{a}\right]^{2}=1$, which simplifies to $x^{2}+y^{2}$ $\frac{2 u}{1+a} x=0$. Therefore, the locus of point $P$ is a circle, but excluding the two intersection points of this circle with the $x$-axis.