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, 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. Substituting the line $\left\{\begin{array}{l}x=t \cos \alpha \\ y=t \sin \alpha\end{array}\right.$ into the equation $(x-4)^{2}+y^{2}=4$ yields $t^{2}-8 t \cos \alpha+12=0$. According to the problem, $\Delta=64 \cos ^{2} \alpha-48=0, \therefore \quad \cos \alpha= \pm \frac{\sqrt{3}}{2}$, and since $0 \leqslant \alpha<\pi$, we have $\alpha=\frac{\pi}{6}$ or $\frac{5 \pi}{6}$, so the answer is A.