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

Center of circle is $\left( 2,-3 \right)$ and radius is $r=10$
Consider the standard equation of the circle is: ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ (equation - 1) And equation of the circle is ${{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=100$ (equation - 2) Now compare both the equations. \begin{align} & {{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=100 \\ & {{\left( x-\left( 2 \right) \right)}^{2}}+{{\left( y-\left( -3 \right) \right)}^{2}}={{\left( 10 \right)}^{2}} \\ \end{align} Center coordinate of circle is $\left( h=2,k=-3 \right)$. And radius of circle is $r=10$. To graph, we plot the points $\left( 2,7 \right)$, $\left( 2,-13 \right)$, $\left( -8,-3 \right)$, and $\left( 12,-3 \right)$ which are, respectively, 10 units above, below, left and right of $\left( 2,-3 \right)$.