$\triangle \mathrm{ABC}$ has vertices $\mathrm{A}(-\mathrm{a}, 0)$ and $\mathrm{B}(\mathrm{b}, 0) \quad(\mathrm{a}>0, \mathrm{b}>0)$ fixed, the third vertex is $\mathrm{C}(\mathrm{x}, \mathrm{y})$, if $\mathrm{CO}(\mathrm{O}$ is the origin) is the bisector of $\angle \mathrm{ACB}$, find the equation of the locus of point $\mathrm{C}$.
Given $\angle \mathrm{ACO}=\angle \mathrm{BOC}$, we have $|\mathrm{CA}|:|\mathrm{CB}| = |A O|:|O B|$, i.e., $\frac{\sqrt{(x+a)^{2}+y^{2}}}{\sqrt{(x-b)^{2}+y^{2}}}=\frac{a}{b}$. Squaring both sides and simplifying, we get $(a-b) x^{2}+(a-b) y^{2}-2 a b x=0$.
(i) When $a=b$, the equation of the locus of point $C$ is $x=0$,
(ii) When $a \neq b$, the equation of the locus of point $C$ is:
$$
x^{2}+y^{2}-\frac{2 a b}{a-b} x=0 \text{. }
$$