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