Answer
(a) Center of circle is $\left( 0,-7555.9 \right)$.
(b) The radius for edge of snowboard is $r=7579\text{mm}$
Work Step by Step
(a) For the center of the circle,
Use the distance formula,
$\begin{align}
& \sqrt{{{\left( -590-0 \right)}^{2}}+{{\left( 0-k \right)}^{2}}}=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( 23-k \right)}^{2}}} \\
& \sqrt{\left( 348100+{{k}^{2}} \right)}=\sqrt{\left( 529+{{k}^{2}}-46k \right)}
\end{align}$
Squaring on both the sides of $\sqrt{\left( 348100+{{k}^{2}} \right)}=\sqrt{\left( 529+{{k}^{2}}-46k \right)}$.
$\begin{align}
& 348100+{{k}^{2}}=529+{{k}^{2}}-46k \\
& 348100-529={{k}^{2}}-{{k}^{2}}-46k \\
& 347571=-46k \\
& \frac{347571}{46}=-k
\end{align}$
$k\approx -7555.9$
Thus, the center of circle is $\left( 0,-7555.9 \right)$.
(b) For the radius,
We use distance formula between center of circle $\left( 0,-7555.9 \right)$ to the any point on board $\left( -590,0 \right),\left( 0,23 \right),\text{ and }\left( 590,0 \right)$. We use $\left( 0,23 \right)$.
Put center of circle $\left( 0,-7555.9 \right)$ and point $\left( 0,23 \right)$ in standard equation of a circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
$\begin{align}
& r=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( -7555.9-23 \right)}^{2}}} \\
& r=7579\text{mm} \\
\end{align}$
Thus, the radius used for edge of snowboard is $r=7579\text{mm}$.
