Answer
The polar coordinates of $\left( 0,-6 \right)$ are $\left( 6,\frac{3\pi }{2} \right)$.
Work Step by Step
The polar coordinates of the point are $\left( r,\theta \right)$.
Now, rewrite the polar coordinates in terms of rectangular coordinates as below:
$r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1)
$\tan \theta =\frac{y}{x}$ …… (2)
Substituting the values of $x\ \text{ and }\ y$ in (1) and (2), we get
$\begin{align}
& r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& =\sqrt{{{\left( 0 \right)}^{2}}+{{\left( -6 \right)}^{2}}} \\
& =\sqrt{36} \\
& r=6
\end{align}$
And,
$\begin{align}
& \tan \theta =\frac{y}{x} \\
& =\frac{-6}{0} \\
& \tan \theta =\infty \left( \text{undefined} \right)
\end{align}$
Hence,
$\tan \theta =\infty $
And,
$\tan \frac{\pi }{2}=\infty $
The $\theta $ lies on the negative y-axis which means
$\begin{align}
& \theta =\pi +\frac{\pi }{2} \\
& =\frac{3\pi }{2}
\end{align}$
Therefore, the polar coordinates of $\left( 0,-6 \right)$ are $\left( 6,\frac{3\pi }{2} \right)$.
