Answer
Circle with equation $x^2+z^2=4$ in the $xz$ plane.
Work Step by Step
Here, the points are the intersection between the equation of the cylinder $x^2+z^2=4$ and the plane $y=0$.
The set of points that satisfies such equations has the equation of a circle lying in the $xz$ plane, centered at origin $O$, having a circle of radius $2$ .
