# Intersection of a Cone and Sphere

Show that a the cone $xy + yz + xz = 0$ cuts the sphere $x^2 + y^2 + z^2 = r^2$ into two equal circles and find their area.

I have been trying to substitute one of the variables, say $z$, from the equation of the cone and putting that into the sphere; this looks like the wrong approach though. Could someone please help me with this?

-

I would start with a change of variables (orthogonal transformation) to diagonalize the quadratic form $xy + yz + xz$. Try $u = (x+y+z)/\sqrt{3}$, $v = (x - y)/\sqrt{2}$, $w = (x + y - 2 z)/\sqrt{6}$.

-
Amazing! That's exactly the new choice of coordinates that I was about to recommend. – Lubin Sep 10 '12 at 20:07
Hi, Thanks a lot for your comment, I will try this approach too. – coolbootgeek Sep 10 '12 at 20:26

The cone's apex is at the origin, so it definitely intersects the origin-centered sphere in equal circles.

Note that the cone contains the coordinate axes. (Simply set any two variables to zero, and see that the last can be arbitrary.) Consequently, it meets the sphere at the points $(\pm r, 0, 0)$, $(0, \pm r, 0)$, $(0,0,\pm r)$, so that the circles of intersection must be circumcircles of equilateral triangles with side-length $r\sqrt{2}$. So ...

-
Great, thanks for the answer – coolbootgeek Sep 10 '12 at 20:25