Answer
Center of circle is $\left( -3,-2 \right)$ and radius is $r=1$
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 ${{x}^{2}}+{{y}^{2}}+6x+4y=-12$ (equation - 2)
Now add $9$ and $4$ on both the sides of equation $\left( 2 \right)$ to complete the square twice.
$\begin{align}
& {{x}^{2}}+{{y}^{2}}+6x+4y+9+4=-12+9+4 \\
& \left( {{x}^{2}}+6x+9 \right)+\left( {{y}^{2}}+4y+4 \right)=1 \\
& {{\left( x+3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{\left( 1 \right)}^{2}}
\end{align}$
Now compare the (equation – 1) with the equation ${{\left( x+3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{\left( 1 \right)}^{2}}$.
Center coordinate of circle is $\left( h=-3,k=-2 \right)$.
And radius of circle is $r=1$.
To graph, we plot the points $\left( -3,-1 \right)$, $\left( -3,-3 \right)$, $\left( -4,-2 \right)$, and $\left( -2,-2 \right)$ which are, respectively, $1$ unit above, below, left and right of $\left( -3,-2 \right)$.
