#### Answer

Two pairs of polar coordinates for this point are:
$(2\sqrt{2}, 225^{\circ})$
$(-2\sqrt{2}, 45^{\circ})$

#### Work Step by Step

$(-2,-2)$
$r = \sqrt{(-2)^2+(-2)^2} = 2\sqrt{2}$
We can find the angle $\phi$ below the negative x-axis:
$tan~\phi = \frac{2}{2}$
$\phi = arctan(\frac{2}{2})$
$\phi = 45^{\circ}$
Then $\theta = 180^{\circ} +45^{\circ} = 225^{\circ}$
Two pairs of polar coordinates for this point are:
$(2\sqrt{2}, 225^{\circ})$
$(-2\sqrt{2}, 45^{\circ})$