I have a maths question where the equation of a sphere $S$ is said to be $x^2+y^2+z^2-12x-6y-4z = 0$, and I'm asked to show that the centre is outside the pyramid whose vertices are $A(12,0,0)$, $B (0,6,0)$, $C (0,0,4)$ and the origin. Edit: These 4 points are also contained within the sphere.
I know how to find the origin and radius etc, but I've spent a good hour on this question and I still don't see how you can do it.
By completing the square, we find that the sphere has standard form $(x - 6)^2 + (y-3)^2 + (z-2)^2 = 49$, so it has centre $(6, 3, 2)$ and radius $7$.
The face of the tetrahedron containing the vertices $A$, $B$, and $C$ is a subset of the plane $x + 2y + 3z = 12$. The plane splits $\mathbb{R}^3$ into two regions which are defined by $x + 2y + 3z \leq 12$ and $x + 2y + 3z > 12$.
As $0 + 2(0) + 3(0) = 0 \leq 12$, the origin is contained in the region defined by $x + 2y + 3z \leq 12$. As this region contains all four vertices of the tetrahedron it must contain the entire tetrahedron.
As $3(6) + 2(3) + 3(2) = 30 > 12$, the centre of the sphere is not contained in the region defined by $x + 2y + 3z \leq 12$. In particular, it is not contained in the tetrahedron.