12. $A, B$ are two given points, $C$ moves on a circle with center $A$, the angle bisector of $\triangle A B C$ at angle $A$ intersects side $B C$ at point $P$, find the locus of $P$.
12. Taking $A$ as the origin and $A B$ as the $x$-axis to establish a rectangular coordinate system. Suppose $C$ moves on the unit circle, the coordinates of point $B$ are $(a, 0)$, the coordinates of point $C$ are $\left(x_{0}, y_{0}\right)$, and the coordinates of point $P$ are $(x, y)$, then
$$
\left\{\begin{array}{l}
x_{0}=\frac{a-(1+a) x}{1-(1+a)}, \\
y_{0}=\frac{-(1+a) y}{1-(1+a)} .
\end{array}\right.
$$

Therefore, $\left[\frac{(1+a) x-a}{a}\right]^{2}+\left[\frac{(1+a) y}{a}\right]^{2}=1$.

That is, $x^{2}+y^{2}-\frac{2 a}{1+a} x=0$.
Thus, the trajectory of $P$ is a circle, but the two points on the $x$-axis should be excluded.