How to see a plane is tangent to a sphere from their equations

Question

Say you have two equations with three variables, the first is the equation of the surface of a sphere and the second of a plane. In this case they intersect in a point $(1,0,0)$. The only way I know to find this point is to rewrite the equation of the sphere so you know its center point and intersect a line going through that point at and at an angle of 90 degrees with the plane. Are there other methods to solve this? Without geometry?

Here are two example equations.

$$\begin{cases} x^2 + y^2 + z^2 - 6x + 6y - 12z + 5&=&0\\ 2x - 3y + 6 z - 2&=&0\\ \end{cases}$$

and the solution

$$\begin{cases} x = 1\\ y = 0\\ z = 0\\ \end{cases}$$

Answer 1

Let us solve the general case. Any sphere $S$ is characterized by an equation $s(x,y,z)=0$, where

$$ s(x,y,z)=x^2+y^2+z^2-2ax-2by-2cz+d, $$

for some parameters $(a,b,c,d)$ such that $d\lt a^2+b^2+c^2$. Likewise, any plane $P$ is characterized by an equation $p(x,y,z)=0$, where

$$ p(x,y,z)=ux+vy+wz+t, $$

for some parameters $(u,v,w,t)$ such that $(u,v,w)\ne(0,0,0)$. That $P$ is tangent to $S$ is equivalent to the condition that a point $(x,y,z)$ belongs to both $P$ and $S$, such that the line between the center $(a,b,c)$ of $S$ and the point $(x,y,z)$ is orthogonal to $P$.

The first part of the condition reads $p(x,y,z)=s(x,y,z)=0$. The vector $(u,v,w)$ is orthogonal to $P$ hence the second part of the condition is that $(x-a,y-b,z-c)$ and $(u,v,w)$ are proportional.

Thus $(x,y,z)=(a+\lambda u,b+\lambda v,c+\lambda w)$ for some $\lambda$. Then $p(x,y,z)=0$ if and only if $$ (u^2+v^2+w^2)\lambda=ua+bv+cw+t, $$ and $s(x,y,z)=0$ if and only if $$ (u^2+v^2+w^2)\lambda^2=a^2+b^2+c^2-d. $$ Thus the plane $P$ is tangent to the sphere $S$ if and only if these two equations have a common solution $\lambda$, which happens if and only if

$$ (ua+bv+cw+t)^2=(u^2+v^2+w^2)\cdot(a^2+b^2+c^2-d). $$

Answer 2

First, a sphere and plane can intersect in a circle, a point, or not at all. As André Nicolas has said, $(1,0,0)$ does not satisfy your equations. For your equations, Alpha shows that the intersection is a circle, not a point.

With the new equation, Alpha still thinks there are more than one point of intersection. The first is now $(x-3)^2+(y+3)^2+(z-6)^2=7^2$ so $(1,0,0)$ is an intersection.