## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

(a) Center of the circle is $\left( 0,-3 \right)$ (b) $r=5\text{ ft}$
(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 $r=5\text{ ft}$.