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. B.

Notice that $a=2, b=\sqrt{m}$,
$$
c=\sqrt{4+m}, e=\frac{\sqrt{4+m}}{2} \text {. }
$$

The equations of the asymptotes are $\frac{x}{2} \pm \frac{y}{\sqrt{m}}=0$, i.e.,
$$
\sqrt{m} x \pm 2 y=0 \text {. }
$$

The distance from the right focus $(\sqrt{4+m}, 0)$ to the asymptote is $d=b=\sqrt{m}$.
Thus, $\sqrt{m}=\frac{\sqrt{4+m}}{2} \Rightarrow m=\frac{4}{3}$.