(4) For the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{m}=1$, the circle with the eccentricity as the radius and the right focus as the center is tangent to the asymptotes of the hyperbola, then $m=(\quad$.
(A) $\frac{3}{2}$
(B) $\frac{4}{3}$
(C) $\frac{5}{4}$
(D) $\frac{6}{5}$
(4) It is easy to know that $a=2, b=\sqrt{m}, c=\sqrt{4+m}, e=\frac{\sqrt{4+m}}{2}$. The equations of the asymptotes: $\frac{x}{2} \pm \frac{y}{\sqrt{m}}=0$, i.e., $\sqrt{m} x \pm 2 y=0$. The distance $d=b=\sqrt{m}$ from the right focus $(\sqrt{4+m}, 0)$ to the asymptote. Thus, we get the equation $\sqrt{m}=\frac{\sqrt{4+m}}{2}$, solving which yields $m=\frac{4}{3}$. Therefore, the answer is B.