# Analytic geometry section of cone and sphere

How to show that the cone $yz+zx+xy=0$ cuts the sphere $x^2+y^2+z^2=a^2$ in two equal circle ?

I understand that the two equations taken together represent the circle. but how to go about finding the equation of these circles ??

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Let's think about this from the geometric point of view first. Isn't it strange that two curved surfaces intersect by a couple of circles? After all, circles are flat, they are essentially planar, and yet we have two three-dimensional surfaces intersect by two circles. So, to get to the bottom of this, I would try and determine the equations of these planes.

This can be done by purely algebraic manipulations. Really, multiply the first equation by $2$ and add it to the second one: $$2(yz+zx+xy) + x^2+y^2+z^2 = a^2.$$ This is the same as: $$(x + y + z)^2 = a^2,$$ which means that if $(x,y,z)$ belongs to our intersection, then either $x+y+z=a$ or $x+y+z=-a$. There we have it: our intersection lies entirely in the union of these two parallel planes: $x+y+z=a$ and $x+y+z=-a$.

So basically instead of intersecting a sphere by a cone we could have intersected it with a pair of planes. Clearly, when we intersect a sphere and a pair of planes, we get a pair of circles.

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That was as amazing an explanation as it could get. Thanks a lot.. I hope you wont mind looking at this one for me math.stackexchange.com/questions/377134/… –  Aman Mittal Apr 30 '13 at 11:32

$(x+y+z)^2=x^2+y^2+z^2+2(yz+zx+xy)=a^2+2\cdot0=a^2$, so the two circles will the sphere intersecting the planes $x+y+z=\pm a$.

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Thanks for this result –  Aman Mittal Apr 30 '13 at 11:33
@AmanMittal, you're welcome! –  Easy Apr 30 '13 at 11:33
Can you please look at this one for me math.stackexchange.com/questions/377134/… –  Aman Mittal Apr 30 '13 at 11:35