Answer
$r=7169\text{mm}$.
Work Step by Step
Use distance formula,
$\begin{align}
& \sqrt{{{\left( -580-0 \right)}^{2}}+{{\left( 0-k \right)}^{2}}}=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( 23.5-k \right)}^{2}}} \\
& \sqrt{\left( 336400+{{k}^{2}} \right)}=\sqrt{\left( 552.25+{{k}^{2}}-47k \right)}
\end{align}$
Squaring on both the sides of $\sqrt{\left( 336400+{{k}^{2}} \right)}=\sqrt{\left( 552.25+{{k}^{2}}-47k \right)}$
$\begin{align}
& 336400+{{k}^{2}}=552.25+{{k}^{2}}-47k \\
& 336400-552.25={{k}^{2}}-{{k}^{2}}-47k \\
& 335847.7=-47k \\
& \frac{335847.7}{47}=-k
\end{align}$
$k\approx -7145.7$
Center of the circle is $\left( 0,-7145.7 \right)$.
For the radius, we use the distance formula between center of circle $\left( 0,-7145.7 \right)$ to the any point on board $\left( -580,0 \right),\left( 0,23.5 \right),\text{ and }\left( 580,0 \right)$
We use $\left( 0,23.5 \right)$
Put the center of circle $\left( 0,-7145.7 \right)$ and point $\left( 0,23.5 \right)$ in the standard equation of a circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
$\begin{align}
& r=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( -7145.7-23.5 \right)}^{2}}} \\
& r=7169\text{mm} \\
\end{align}$
Thus, the radius used for the edge of the snowboard is $r=7169\text{mm}$.
