To convert the point $M(6, -6\sqrt{3})$ from rectangular (Cartesian) coordinates to polar coordinates $(r, \theta)$, we use the following formulas:

1. Calculate the radius $r$: $r = \sqrt{x^2 + y^2}$ where $x = 6$ and $y = -6\sqrt{3}$.

   $r = \sqrt{6^2 + (-6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12$

2. Calculate the angle $\theta$: $\theta = \tan^{-1}\left(\frac{y}{x}\right)$ $\theta = \tan^{-1}\left(\frac{-6\sqrt{3}}{6}\right) = \tan^{-1}(-\sqrt{3})$

   The angle whose tangent is $-\sqrt{3}$ is $\theta = -\frac{\pi}{3}$.

3. Determine the correct quadrant: The point $M(6, -6\sqrt{3})$ is in the fourth quadrant because $x > 0$ and $y < 0$. The angle $\theta = -\frac{\pi}{3}$ correctly places the point in the fourth quadrant.

Therefore, the polar coordinates of the point $M(6, -6\sqrt{3})$ are: $\boxed{(12, -\frac{\pi}{3})}$ where $r = 12$ and $\theta = -\frac{\pi}{3}$.

Do you want further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would you convert the polar coordinates $(12, -\frac{\pi}{3})$ back to rectangular coordinates?
2. What would be the polar form of the point if it were located in the second quadrant?
3. How does the sign of the angle $\theta$ affect the location of the point on the polar coordinate plane?
4. What are the possible values of $\theta$ for the same point $M$ in different quadrants?
5. How do polar coordinates help in understanding the symmetry of points on the coordinate plane?

Tip: When converting between rectangular and polar coordinates, always check the quadrant of the point to ensure that the angle $\theta$ is correctly placed.