Answer
See the graph below:
Work Step by Step
We know that a circle has center $\left( h,k \right)$ and radius $r$; the general equation of the circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
Thus, the given equation ${{x}^{2}}+{{y}^{2}}+4x-6y-3=0$ can be rewritten as
$\begin{align}
& {{x}^{2}}+{{y}^{2}}+4x-6y-3=0 \\
& \left( {{x}^{2}}+4x \right)+\left( {{y}^{2}}-6y \right)=3 \\
& {{x}^{2}}+2\left( 2 \right)\left( x \right)+{{2}^{2}}+{{y}^{2}}-2\left( 3 \right)\left( y \right)+{{3}^{2}}=3+{{2}^{2}}+{{3}^{2}} \\
& {{\left( x+2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}=16
\end{align}$
${{\left( x+2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{4}^{2}}$
So, the center of the circle is $\left( h,k \right)=\left( -2,3 \right)$ and radius $r=4$.
Now, in order to plot this circle on the rectangular coordinate system, take $\left( -2,3 \right)$ as a center and draw a circle with radius $r=4$.
It can be observed that for the x-axis the graph is going from $\left[ -6,2 \right]$. So, the domain of the graph is $\left[ -6,2 \right]$.
Also, it can be seen that for the y-axis the graph is going from $\left[ -1,7 \right]$. So, the range of the graph is $\left[ -1,7 \right]$.
