Answer
$(x+4)^2+(y+2)^2=4$
$C(-4,-2), r=2$
See graph.
Work Step by Step
Step 1. From the given equation $x^2+y^2+8x+4y+16=0$, we have $(x^2+8x+16)+(y^2+4y+4)=16+4-16$ and $(x+4)^2+(y+2)^2=4$
Step 2. We can identify the center and radius of the circle as $C(-4,-2), r=2$
Step 3. See graph.
