The polar co-ordinates of the point whose cartesian co-ordinates are $(-2,-2)$, are given by
The polar co-ordinates of the point whose cartesian co-ordinates are $(-2,-2)$, are given by
$\left(2 \sqrt{2}, \frac{5 \pi}{4}\right)$
$\left(2 \sqrt{2}, \frac{3 \pi}{4}\right)$
$\left(2 \sqrt{2}, \frac{7 \pi}{6}\right)$
$\left(2 \sqrt{2}, \frac{\pi}{4}\right)$
Solution
(C)
We know that $r=\sqrt{x^{2}+y^{2}}=\sqrt{(-2)^{2}+(-2)^{2}}=2 \sqrt{2}$ and $\tan \theta=\frac{y}{x}=\frac{-2}{-2}=1$ $\theta=\tan ^{-1} 1=\left(\pi+\frac{\pi}{4}\right)=\frac{5 \pi}{4}$, as the point $(-2,-2)$ lies in III quadrant. $\therefore(r, \theta)=\left(2 \sqrt{2}, \frac{5 \pi}{4}\right)$