Answer
$(x+3)^2+(y+1)^2=4$
$C(-3,-1), r=2$
See graph.
Work Step by Step
Step 1. From the given equation $x^2+y^2+6x+2y+6=0$, we have $(x^2+6x+9)+(y^2+2y+1)=9+1-6$ and $(x+3)^2+(y+1)^2=4$
Step 2. We can identify the center and radius of the circle as $C(-3,-1), r=2$
Step 3. See graph.
