Answer
Center of circle is $\left( -3,2 \right)$ and radius is $r=\sqrt{28}$
Work Step by Step
Standard equation of the circle is:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ (equation -1)
And equation of circle is ${{x}^{2}}+{{y}^{2}}+6x-4y=15$ (equation -2)
Now add $4$ and $9$ on both the sides of (equation -2) to complete the square twice.
$\begin{align}
& {{x}^{2}}+{{y}^{2}}+6x-4y+9+4=15+9+4 \\
& \left( {{x}^{2}}+6x+9 \right)+\left( {{y}^{2}}-4y+4 \right)=28 \\
& {{\left( x+3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( \sqrt{28} \right)}^{2}}
\end{align}$
Now compare the standard equation with the equation${{\left( x+3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( \sqrt{28} \right)}^{2}}$.
Center coordinate of circle is $\left( h=-3,k=2 \right)$.
And radius of circle is $r=\sqrt{28}$.
To graph, we plot the points $\left( -3,7.292 \right)$, $\left( -3,-3.292 \right)$, $\left( -8.292,2 \right)$, and $\left( 2.292,3 \right)$ which are, respectively, $\sqrt{28}$ units above, below, left and right of $\left( -3,2 \right)$.
