Example 1 As shown in Figure $11-1$, given a fixed point $A(a, 0),(a>0)$ and a line $l: x=-1, B$ is a moving point on line $l$, the angle bisector of $\angle B(A$ intersects $A B$ at point $C$. Find the equation of the locus of point $C$.
(1999 National College Entrance Examination Question)
Let $B(-1, b)$, then the equations of lines $O A$ and $O B$ are $y=0$ and $y=-b x$, respectively. Let $C(x, y)$, then $0 \leqslant x<a$. Since $O C$ bisects $\angle A O B$, the distances from $C$ to $O A$ and $O B$ are equal. Using the point-to-line distance formula, we have $|y|=\frac{|y+b x|}{\sqrt{1+b^{2}}}$.
Also, $C$ lies on line $A B$, so $y=-\frac{b}{1+a}(x-a)$.
From (1) and (2), eliminating $b$, simplifying, and rearranging, we get $y^{2}\left[(1-a) x^{2}-2 a x+(1+a) y^{2}\right]=0$.
If $y \neq 0$, then $(1-a) x^{2}-2 a x+(1+a) y^{2}=0(0<x<a)$;
If $y=0$, then $b=0, \angle A(B B=\pi$, and the coordinates of point $C$ are $(0,0)$, which also satisfies the above equation. In summary, the equation of the locus of point $C$ is $(1-a) x^{2}-2 a x+(1+a) y^{2}=0(0 \leqslant x<a)$.