Answer
circle $x^2+y^2=16$ in $z=0$
Work Step by Step
Since we have two equations that shows the intersection between the equation of sphere $x^2+y^2+(z+3)^2=25$ and a plane $z=0$.
Re-write the equation of sphere as: $x^2+y^2+(0+3)^2=25 $
or, $x^2+y^2=16$
Thus, we have the set of points that satisfies the equation of a circle $x^2+y^2=16$ in $z=0$ center: $(0,0,0)$ with radius of $4$ .
