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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 $\pm \sqrt3$.

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$2$ 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.

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 $60 \deg$.enter image description here

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

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

enter image description here

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:

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

Aligator
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  • I edited answer and write the equation – Aligator Jun 28 '20 at 12:17
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    If $x^2+ y^2= 1$ and $x^2+ y^2+ z^2= 4$ then $1+ z^2= 4$ so $z^2= 3$ and $z= \pm\sqrt{3}$. That is, the cylinder (not circle) intersects the sphere (as Soheil said) where $x^2+ y^2= 1$ and $z= \pm\sqrt{3}$. We can parameterize those two circles as $x= cos(t)$, $y= sin(t)$, $z= \sqrt{3}$ and $x= cos(t)$, $y= sin(t)$, and $z= -\sqrt{3}$. – user247327 Jun 28 '20 at 12:32