9 As shown in the figure, construct an equilateral triangle $P C D$. Since $\triangle A B C$ is also an equilateral triangle, it can be proven that
$$
\triangle A C P \cong \triangle B C D,
$$

Therefore, $B D=A P$. Also, since
$$
B D=P B+P C=P B+P D,
$$

points $B, P, D$ are collinear. Because $\angle C B P=\angle P A C$, point $P$ lies on the circumcircle of $\triangle A B C$. Since $P A>P B, P A>P C$, the locus of point $P$ is given by the equation
$$
x^{2}+(y-1)^{2}=4(y \leqslant 0) .
$$
9. Given three points on the coordinate plane $A(0,3)、B(-\sqrt{3}, 0)、C(\sqrt{3}, 0)$, and $P$ is a point on the coordinate plane such that $P A=P B+P C$, then the equation of the locus of point $P$ is