Step 1

To convert from polar to Cartesian from.

Step 2

Given:

\(\displaystyle{r}={5}{\sin{\theta}}\)

Step 3

Formula used,

\(\displaystyle{x}={r}{\cos{\theta}}{\quad\text{and}\quad}{y}={r}{\sin{{0}}}\)

\(\displaystyle{r}^{{{2}}}={x}^{{{2}}}+{y}^{{{2}}}\)

Step 4

Simplify as,

\(\displaystyle{r}={5}{\sin{\theta}}\)

Multiply by r as,

\(\displaystyle{r}^{{{2}}}={5}{r}{\sin{\theta}}\)

\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={5}{\left({r}{\sin{\theta}}\right)}\ \ \ {\left({sin{{c}}}{e},{r}^{{{2}}}={x}^{{{2}}}+{y}^{{{2}}}\right)}\)

\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={5}{y}\ \ \ {\left({sin{{c}}}{e},{y}={\left({r}{\sin{\theta}}\right)}\right)}\)

\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}-{5}{y}={0}\)

\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}-{5}{y}+{\frac{{{25}}}{{{4}}}}={\frac{{{25}}}{{{4}}}}\)

\(\displaystyle{x}^{{{2}}}+{\left({\frac{{{y}-{5}}}{{{2}}}}\right)}^{{{2}}}={\left({\frac{{{5}}}{{{2}}}}\right)}^{{{2}}}\)

Step 5

Hence, the Cartesian form of the equation is \(\displaystyle{x}^{{{2}}}+{\left({\frac{{{y}-{5}}}{{{2}}}}\right)}^{{{2}}}={\left({\frac{{{5}}}{{{2}}}}\right)}^{{{2}}}\) which is the equation of circle with center \(\displaystyle{\left({0},{\frac{{{5}}}{{{2}}}}\right)}\) and radius \(\displaystyle{\frac{{{5}}}{{{2}}}}\).