4. Given the line $\left\{\begin{array}{l}x=t \cos \alpha, \\ y=t \sin \alpha\end{array}\right.$ ( $t$ is a parameter) and the circle $\left\{\begin{array}{l}x=4+2 \cos \theta, \\ y=2 \sin \theta\end{array}\right.$ ( $\theta$ is a parameter) are tangent to each other. Then the inclination angle of the line is ( ).
(A) $\frac{\pi}{6}$ or $\frac{5 \pi}{6}$
(B) $\frac{\pi}{4}$ or $\frac{3 \pi}{4}$
(C) $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$
(D) $-\frac{\pi}{6}$ or $-\frac{5 \pi}{6}$
4. (A).

Substitute the line $\left\{\begin{array}{l}x=t \cos \alpha, \\ y=t \sin \alpha\end{array}\right.$ into the equation $(x-4)^{2}+y^{2}=4$ to get
$$
t^{2}-8 t \cos \alpha+12=0 \text {. }
$$

By the problem, $\Delta=64 \cos ^{2} \alpha-48=0$,
$$
\therefore \cos \alpha= \pm \frac{\sqrt{3}}{2} \text {. }
$$

Also, $0 \leqslant \alpha<\pi$, thus, $\alpha=\frac{\pi}{6}$ or $\frac{5 \pi}{6}$.