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