Answer
$(2\sqrt{2}, \displaystyle \frac{7\pi}{6})$
Work Step by Step
To change from rectangular to polar coordinates, use
$r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$.
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$r^{2}=(-\sqrt{6})^{2}+(-\sqrt{2})^{2}=6+2=8$
$r=\pm 2\sqrt{2}$
$\displaystyle \tan^{-1}(\frac{-\sqrt{2}}{-\sqrt{6}})=\tan^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi}{6}$, which is in quadrant I.
The point P$(-\sqrt{6},-\sqrt{2})$ is in quadrant III,
so we take $\theta= \displaystyle \frac{\pi}{6}+\pi=\frac{7\pi}{6}$.
The terminal end of $\theta$ passes through P, so for this angle, r is positive.
Polar coordinates: $(2\sqrt{2}, \displaystyle \frac{7\pi}{6})$
