# Replace the polar equations with equivalent Cartesian equations. Then describe or identify the graph. r^{2} = -6r sin theta

Replace the polar equations with equivalent Cartesian equations. Then describe or identify the graph. $$\displaystyle{r}^{{{2}}}=-{6}{r}{\sin{\theta}}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

comentezq

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