Find the geometric position of all points satisfying the two equations $x^2+y^2+z^2=4$ and $x^2+y^2=1$

Question

Find the geometric position of all points satisfying the two equations $x^2+y^2+z^2=4$ and $x^2+y^2=1$

I think the points for which their coordinates satisfies the two equations at the same time,are all on a circle $x^2+y^2=1$ such that the distance between the circle and the origin is $\pm \sqrt3$.

Answer 1

$x^2+y^2+z^2=4$ represent a sphere with origin $O$ and the $r=2$

$x^2+y^2=1$ represent a cylinder with Orgin $O$ and the $r=1$

If the center of the sphere lies on the axis of the cylinder of revolution, then the intersection curve degenerates into two circles.

The equation of intersection( 2 circles) is:

$x^2+y^2=1 , z=\pm \sqrt3$

Answer 2

$2$ circles in fact, and the parameterized equations are $f(t) = (t, \pm \sqrt{1 - t^2}, \sqrt{3})$ and $g(t) = (t, \pm \sqrt{1 - t^2}, -\sqrt{3})$. So you are pretty close.

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Then you can see these 2 circles are actually the parallels (horizontal strips on any sphere) at $60 \deg$ north and south from the equator of the sphere with radius of $2$ units and center at the origin $(0, 0, 0)$.

See my finger illustration for why $60 \deg$.