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

ringearV 2021-09-07 Answered
Replace the polar equations with equivalent Cartesian equations. Then describe or identify the graph. \(\displaystyle{r}^{{{2}}}=-{6}{r}{\sin{\theta}}\)

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comentezq
Answered 2021-09-08 Author has 20779 answers

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

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