Answer
(a) Center of the circle is $\left( 0,-1727.25 \right)$
(b) $r=1728.75\text{ cm}$
Work Step by Step
(a) For the center of the circle,
Use distance formula,
$\begin{align}
& \sqrt{{{\left( 72-0 \right)}^{2}}+{{\left( 0-k \right)}^{2}}}=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( 1.5-k \right)}^{2}}} \\
& \sqrt{\left( 5184+{{k}^{2}} \right)}=\sqrt{\left( 2.25+{{k}^{2}}-3k \right)}
\end{align}$
Squaring on both the sides of $\sqrt{\left( 5184+{{k}^{2}} \right)}=\sqrt{\left( 2.25+{{k}^{2}}-3k \right)}$.
$\begin{align}
& 5184+{{k}^{2}}=2.25+{{k}^{2}}-3k \\
& 5184-2.25={{k}^{2}}-{{k}^{2}}-3k \\
& 5181.75=-3k \\
& \frac{5181.75}{3}=-k
\end{align}$
Simplify More,
$k\approx -1727.25$
Thus, the center of the circle is $\left( 0,-1727.25 \right)$
(b) For the radius,
We will use the distance formula between center of circle $\left( 0,-1727.25 \right)$ to the any point on board $\left( 72,0 \right)\text{ and }\left( 0,1.5 \right)$
We will use$\left( 0,1.5 \right)$.
Put center of circle $\left( 0,-1727.25 \right)$ and point $\left( 0,1.5 \right)$ in the standard equation of a circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
$\begin{align}
& r=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( -1727.25-1.5 \right)}^{2}}} \\
& r=1728.75 \\
\end{align}$
Thus, the radius $r=1728.75\text{ cm}$.
