4. Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0), l$ is an asymptote of the hyperbola $C$, and $F_{1}, F_{2}$ are the left and right foci of the hyperbola $C$. If the symmetric point of $F_{1}$ with respect to the line $l$ lies on the circle $(x-c)^{2}+y^{2}=c^{2}$ (where $c$ is the semi-focal distance of the hyperbola), then the eccentricity of the hyperbola $C$ is $\qquad$
Given that $O F_{1}=O M=O F_{2}=F_{2} M$, then $\triangle M F_{1} F_{2}$ is a right triangle, and $\triangle M O F_{2}$ is an equilateral triangle. Therefore, the inclination angle of the asymptote of the hyperbola is $120^{\circ}$.
So $\frac{b}{a}=\tan 60^{\circ}=\sqrt{3} \Rightarrow e=\sqrt{1+\frac{b^{2}}{a^{2}}}=2$.