Ans: D
Let $\theta$ be the acute angle between $PO$ and the negative $x$-axis.
Let $\theta$ be the acute angle between $PO$ and the negative $x$-axis.
Then, we have
$\begin{array}{rcl}
\tan \theta & = & \dfrac{-3\sqrt{3}}{-3} \\
\theta & = & 60^\circ
\end{array}$
Also,
$\begin{array}{rcl}
OP & = & \sqrt{(-3-0)^2 + (-3\sqrt{3}-0)^2} \\
& = & 6
\end{array}$
Therefore, the polar coordinates of $P$ are $(6, 240^\circ)$.
Hence, the polar coordinates of the image are $(6, 330^\circ)$.
