Answer
Circle with equation $x^2+y^2=16$, in the $xy$ plane.
Work Step by Step
Here, the the two equations represent the intersection between the equation of a sphere $x^2+y^2+(z+3)^2=25$ and the plane $z=0$.
Now, $x^2+y^2+(0+3)^2=25 \implies x^2+y^2=25-9$
or, $x^2+y^2=16$
The set of points that satisfies the equation $x^2+y^2=16$ is a circle lying in $z=0$, centered at (0,0,0), having a circle of radius $4$ .
