Answer
Center: $(-6,3)$ $;$ Radius: $7$
Work Step by Step
$x^{2}+y^{2}+12x-6y-4=0$
Take $4$ to the right side:
$x^{2}+y^{2}+12x-6y=4$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}+12x)+(y^{2}-6y)=4$
Complete the square for each group by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside each parentheses and also on the right side of the equation.
For the first group, $b=12$ and for the second group, $b=-6$
$\Big[x^{2}+12x+\Big(\dfrac{12}{2}\Big)^{2}\Big]+\Big[y^{2}-6y+\Big(-\dfrac{6}{2}\Big)^{2}\Big]=4+\Big(\dfrac{12}{2}\Big)^{2}+\Big(-\dfrac{6}{2}\Big)^{2}$
$(x^{2}+12x+36)+(y^{2}-6y+9)=4+36+9$
$(x^{2}+12x+36)+(y^{2}-6y+9)=49$
Factor the expressions inside each parentheses, which are perfect square trinomials:
$(x+6)^{2}+(y-3)^{2}=49$
This is now the standard form of the equation of a circle, which is $(x-h)^{2}+(y-k)^{2}=r^{2}$
It can be seen from the obtained equation that $(h,k)=(-6,3)$, so the center of the circle is the point $(-6,3)$
The radius is:
$r^{2}=49$
$\sqrt{r^{2}}=\sqrt{49}$
$r=7$
The graph is:
