Answer
A circle of radius $\sqrt{3}$, centered at the origin, in the xz-plane.
Work Step by Step
$x^{2}+(y-1)^{2}+z^{2}=4$
is a sphere centered at $(0,1,0)$, with radius $2$.
$y=0$ is the xz-plane.
The equation of their intersection in the xz-plane is obtained by substituting $y=0$ into the equation of the sphere:
$x^{2}+1+z^{2}=4$
$x^{2}+z^{2}=3$, in the xz-plane.
A circle of radius $\sqrt{3}$, centered at the origin, in the xz-plane.
