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

Center of the circle is $\left( -4,3 \right)$ and radius is $r=\sqrt{40}$
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 ${{x}^{2}}+{{y}^{2}}+8x-6y=15$ (equation - 2) Now add $16$ and $9$ on both the sides of equation $\left( 2 \right)$ to complete the square twice. \begin{align} & {{x}^{2}}+{{y}^{2}}+8x-6y+16+9=15+16+9 \\ & \left( {{x}^{2}}+8x+16 \right)+\left( {{y}^{2}}-6y+9 \right)=40 \\ & {{\left( x+4 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( \sqrt{40} \right)}^{2}} \end{align} Compare the standard equation with the equation${{\left( x+4 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( \sqrt{40} \right)}^{2}}$. Center coordinate of circle is $\left( h=-4,k=3 \right)$. And radius of circle is $r=\sqrt{40}$. To graph, we plot the points $\left( -4,9.325 \right)$, $\left( -4,-3.325 \right)$, $\left( -10.325,3 \right)$, and $\left( 2.325,3 \right)$ which are, respectively, $\sqrt{40}$units above, below, left and right of $\left( -4,3 \right)$.