Answer
See below:
Work Step by Step
Consider the polar equation $r=4\sin \theta $.
Convert the polar equation into the rectangular equation,
Since, $x=r\cos \theta $, $y=r\sin \theta $ and ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$
Multiply r in both sides of the equation $r=4\sin \theta $ ,
$\begin{align}
& r\left( r \right)=r\left( 4\sin \theta \right) \\
& {{r}^{2}}=4r\sin \theta
\end{align}$
Now, substitute the values $x=r\cos \theta $, $y=r\sin \theta $ and ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ in equation ${{r}^{2}}=4r\sin \theta $ ,
$\begin{align}
& {{r}^{2}}=4r\sin \theta \\
& {{x}^{2}}+{{y}^{2}}=4y \\
\end{align}$
Rewrite the equation ${{x}^{2}}+{{y}^{2}}=4y$ in the form of ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
${{x}^{2}}+{{y}^{2}}=4y$
Subtract $4y$ from both sides of the equation,
$\begin{align}
& {{x}^{2}}+{{y}^{2}}-4y=4y-4y \\
& {{x}^{2}}+{{y}^{2}}-4y=0 \\
\end{align}$
Add ${{\left( 2 \right)}^{2}}$ to both sides of the equation,
${{x}^{2}}+{{y}^{2}}-4y+{{\left( 2 \right)}^{2}}={{\left( 2 \right)}^{2}}$
Now apply the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ ,
$\begin{align}
& {{x}^{2}}+{{y}^{2}}-4y+{{\left( 2 \right)}^{2}}={{\left( 2 \right)}^{2}} \\
& {{x}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( 2 \right)}^{2}} \\
& {{\left( x-0 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( 2 \right)}^{2}} \\
\end{align}$
Now, equate equation ${{x}^{2}}+{{y}^{2}}=4y$ to the equation of the circle, that is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$; here, r is the radius of the circle and $\left( h,k \right)$ is the center point of the circle.
So, the center point of the circle is $\left( 0,2 \right)$ and the radius of the circle is 2.
