Answer
Center: $(-3,-1)$ $;$ Radius: $2$
Work Step by Step
$x^{2}+y^{2}+6x+2y+6=0$
Take $6$ to the right side:
$x^{2}+y^{2}+6x+2y=-6$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}+6x)+(y^{2}+2y)=-6$
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, $b=6$ and for the second group, $b=2$
$\Big[x^{2}+6x+\Big(\dfrac{6}{2}\Big)^{2}\Big]+\Big[y^{2}+2y+\Big(\dfrac{2}{2}\Big)^{2}\Big]=-6+\Big(\dfrac{6}{2}\Big)^{2}+\Big(\dfrac{2}{2}\Big)^{2}$
$(x^{2}+6x+9)+(y^{2}+2y+1)=-6+9+1$
$(x^{2}+6x+9)+(y^{2}+2y+1)=4$
Factor the expressions inside each parentheses, which are perfect square trinomials:
$(x+3)^{2}+(y+1)^{2}=4$
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 $(h,k)=(-3,-1)$, so the center of the circle is the point $(-3,-1)$
The radius is:
$r^{2}=4$
$\sqrt{r^{2}}=\sqrt{4}$
$r=2$
The graph is:
