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