Answer
(a) Center of the circle is $\left( 0,-3 \right)$
(b) $r=5\text{ ft}$
Work Step by Step
(a) For the center of the circle,
Use the distance formula,
$\begin{align}
& \sqrt{{{\left( 0-\left( -4 \right) \right)}^{2}}+{{\left( y-0 \right)}^{2}}}=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( y-2 \right)}^{2}}} \\
& \sqrt{\left( 16+{{y}^{2}} \right)}=\sqrt{\left( 4+{{y}^{2}}-4y \right)}
\end{align}$
Squaring on both the sides of $\sqrt{\left( 16+{{y}^{2}} \right)}=\sqrt{\left( 4+{{y}^{2}}-4y \right)}$.
$\begin{align}
& \left( 16+{{y}^{2}} \right)=\left( 4+{{y}^{2}}-4y \right) \\
& 16-4={{y}^{2}}-{{y}^{2}}-4y \\
& 12=-4y \\
& \frac{12}{4}=-y
\end{align}$
Simplify,
$y\approx -3$
Thus, the center of circle is $\left( 0,-3 \right)$.
(b) For radius,
We will use distance formula between center of circle $\left( 0,-3 \right)$ to the any point on board$\left( -4,0 \right)\text{ and }\left( 0,2 \right)$
We use$\left( 0,2 \right)$.
Put center of circle $\left( 0,-3 \right)$and point $\left( 0,2 \right)$ in standard equation of a circle${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
$\begin{align}
& r=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( -3-2 \right)}^{2}}} \\
& r=5 \\
\end{align}$
Thus, the radius is $r=5\text{ ft}$.
