1. **State the problem:** Convert the Cartesian coordinate $(2, -6)$ to polar coordinates $(r, \theta)$ where $0 \leq \theta < 2\pi$. 2. **Recall the formulas:** - The radius $r$ is given by $$r = \sqrt{x^2 + y^2}$$ - The angle $\theta$ is given by $$\theta = \arctan\left(\frac{y}{x}\right)$$ but must be adjusted based on the quadrant. 3. **Calculate $r$:** $$r = \sqrt{2^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$$ 4. **Calculate $\theta$:** $$\theta = \arctan\left(\frac{-6}{2}\right) = \arctan(-3)$$ 5. **Determine the quadrant:** Since $x=2 > 0$ and $y=-6 < 0$, the point is in the fourth quadrant. 6. **Adjust $\theta$ for the fourth quadrant:** The arctangent of a negative number gives a negative angle in the fourth quadrant, so to get $\theta$ in $[0, 2\pi)$, add $2\pi$: $$\theta = 2\pi + \arctan(-3)$$ 7. **Final polar coordinates:** $$\boxed{\left(2\sqrt{10}, 2\pi + \arctan(-3)\right)}$$ This is the polar form of the point $(2, -6)$ with $r = 2\sqrt{10}$ and $\theta$ between $0$ and $2\pi$.