\(\displaystyle{r}^{{{2}}}=-{6}{r}{\sin{\theta}}\)

The Cartesian equation is \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}=-{6}{y}{\left[\because{x}={r}{\cos{\theta}},{y}={r}{\sin{\theta}}\ \text{and}\ {r}=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}}}\right]}\)

\(\displaystyle\Rightarrow{x}^{{{2}}}+{y}^{{{2}}}+{6}{y}={0}\)

\(\displaystyle\Rightarrow{x}^{{{2}}}+{y}^{{{2}}}+{6}{y}+{9}={9}\)

\(\displaystyle\Rightarrow{x}^{{{2}}}+{\left({y}+{3}\right)}^{{{2}}}={9}\)

\(\displaystyle\Rightarrow{\left({x}-{0}\right)}^{{{2}}}+{\left({y}-{\left(-{3}\right)}\right)}^{{{2}}}={3}^{{{2}}}\)

Now its the form of of\(\displaystyle\Rightarrow{\left({x}-{h}\right)}^{{{2}}}+{\left({y}-{k}\right)}^{{{2}}}={r}^{{{2}}}\)

Hence it represents a circle with centre at \((h,k)(0,-3)\) and radius 3