Answer
A circle, centered at $(-2,0)$, with radius $4.$
Work Step by Step
$(x^{2}+4x)+y^{2}=12\qquad $
We group like terms
$(x^{2}+2\cdot 2x+2^{2})+y^{2}=12+2^{2}\qquad$
We complete the square:
$(x+2)^{2}+y^{2}=16$
This equation is of a circle, centered at $(-2,0)$, with radius $4.$
