3. 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 $\qquad$
3. $x^{2}+y^{2}=4$.

According to the problem, the symmetric point $F_{1}^{\prime}$ of point $F_{1}$ with respect to the line $P M$ lies on the line $P F_{2}$, then $P F_{1}^{\prime}=P F_{1}$: According to the definition of a hyperbola, we get $\left|F_{1}^{\prime} F_{2}\right|$
$$
\begin{array}{l}
=\left\|P F_{1}^{\prime}|-| P F_{2}\right\| \\
=\left\|P F_{1}|-| P F_{2}\right\| \\
=2 a=4 .
\end{array}
$$

Let $M(x, y)$, then
$$
F_{1}^{\prime}(2 x+2 \sqrt{2}, 2 y) \text {. }
$$

Substituting into the above equation, we get
$\sqrt{(2 x)^{2}+(2 y)^{2}}=4$, i.e., $x^{2}+y^{2}=4$.