Answer
Circle with equation $x^2+z^2=4$ in the $xz$ plane.
Work Step by Step
Since, we have the points that are the intersection between the equation of cylinder $x^2+z^2=4$ and the plane $y=0$.
The set of points that satisfies such equations will have the equation of a circle $x^2+z^2=4$ in the $xz$ plane, whose center is at origin $O$ and having circle of radius $2$ .
