4. Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $x^{2}-y^{2}=4$, and $P$ be any point on the hyperbola. A perpendicular line is drawn from $F_{1}$ to the angle bisector of $\angle F_{1} P F_{2}$, with the foot of the perpendicular being $M$. Then the equation of the locus of point $M$ is ( ).
(A) $x^{2}+y^{2}=1$
(B) $x^{2}+y^{2}=2$
(C) $x^{2}+y^{2}=4$
(D) $x^{2}+y^{2}=8$
4. C.

Extend $F_{1} M$ to intersect with $P F_{2}$ (or the extension of $P F_{2}$) at point $N$. Then $\triangle P F_{1} N$ is an isosceles triangle, with $M$ being the midpoint of the base $F_{1} N$. Connect $O M$. Then
$$
O M=\frac{1}{2} F_{2} N=\frac{1}{2}\left|P F_{1}-P F_{2}\right|=2 .
$$

Therefore, the answer is (C).