Ans: C
Let $A’$ be the image. Then the rectangular coordinates of $A’$ are $(-\sqrt{3},-1)$. Hence, we have
$\begin{array}{rcl}
r & = & \sqrt{(-\sqrt{3})^2 + (-1)^2} \\
r & = & 2
\end{array}$
Also,
$\begin{array}{rcl}
\tan \theta & = & \dfrac{-1}{-\sqrt{3}} \\
\theta & = & 210^\circ
\end{array}$
Therefore, the polar coordinates of $A’$ are $(2, 210^\circ)$.
