Find two sets of polar coordinates for the point between $0\leq \theta <2\pi$

Question

The point is $\left(-2\sqrt{3},-2\right)$

Using the formula $x^2 +y^2=r^2$ and $\tan(\frac{y}{x})$, plugging in we get $$(-2\sqrt{3})^2+(-2)^2=r^2$$ $$r=4$$ $$\tan(\frac{-2}{-2\sqrt{3}})=\tan\left(\frac{\sqrt{3}}{3}\right)=\frac{\pi}{6}$$ So my question is, which set of coordinates is correct? $$\left(4, \frac{\pi}{6}\right),\left(-4,\frac{7\pi}{6}\right)$$ or $$\left(-4, \frac{\pi}{6}\right),\left(4,\frac{7\pi}{6}\right)$$

Answer 1

Because the point $\left(-2\sqrt{3},-2\right)$ is in the third quadrant ($x$- and $y$- coordinates are negative),

the correct set of coordinates must be the latter.

Answer 2

Since $x<0$ we have that

$$\theta = \arctan \frac y x + \pi$$

with $r\ge0$ of course, since negative values for $r$ are not allowed, therefore $(r, \theta)=\left(4,\frac{7\pi}{6}\right)$.

Refer also to Converting between polar and Cartesian coordinates.