How to prove that a plane is a tangent plane to a sphere? I know how to find a tangent plane to a sphere at any given point on the sphere.
But I have been given a tangent plane and a sphere, and need to prove, that the plane is indeed a tangent plane to a circle. 
The circle has a radius of 5, center of $(-1,3,0)$ and equation of $(x+1)^2 + (y-3)^2 + z^2 = 25$.
The plane has an equation of $4x+3z+29=0$.
How to I prove that that the plane is a tangent plane first of all, and then how would i find the point at which it touches the circle?
 A: Let $f(x,y,z) = (x+1)^2 + (y-3)^2+z^2$. The gradient of $f$ is normal to the surface of the sphere. Meanwhile the normal to the plane is $(4,0,3)$. So find $x$, $y$, and $z$ values for which $\nabla f$ is a multiple of $(4, 0, 3)$. Then set the multiple so that that these $x$, $y$, and $z$ values lie on the surface of the sphere.
A: Suppose your plane is given in the form x . n = a, where n is a unit vector normal to the plane, the sphere center is located at c and the radius is R.
What you want to check is if the perpendicular distance from c to the plane is R. Suppose p is some point on the plane. We can write c = p + u where u is the vector from p to c. We can do this for any point on the plane, but for some particular point |u| is minimal. In this case u is normal to the plane, i.e. u = $\alpha$ n, where $\alpha$ is a scalar, the (signed) perpendicular distance. We have c = p + $\alpha$ n. 
All we need to do now to find $\alpha$ is to take the inner product of this with n, giving c . n = p . n +$\alpha$ n . n. Note that p . n = a since p belongs to the plane and n . n =1 since n is a unit vector, giving $\alpha$ = c . n - a. The sphere is tangent to the plane if and only if |$\alpha$| = R or equivalently if |c . n - a| = R. 
Having found $\alpha$ we get c = p +(c . n -a) n from which we get  p = c + (a - c . n) n. This is the foot of the perpendicular from the sphere center to the plane which coincides with the point of contact if the sphere is tangent to the plane.
In your case c = (-1,3,0), R=5, n = (4/5,0,3/5) (normalized) and a=-29/5. c . n - a = -4/5+29/5=5=R so yes, the plane is tangent to the sphere. The point of contact is the foot of the perpendicular, p = c + (a - c . n) n = c - 5 n = (-5,3,-3)  Q.E.D.
A: A common way is to solve the section of the plane and the sphere
$$\begin{cases}(x+1)^2+(y−3)^2+z^2=25\\4x+3z+29=0\end{cases}$$
by eliminating z:
$$\frac{25x^2}9+\frac{250x}9+y^2-6y+\frac{706}9=0$$
Then this equation can be written as:
$$\frac{25(x+5)^2}9+(y-3)^2=0$$
So the section projecting to z plane is a point $(-5,3)$, which is the tangent point.
Eliminating x can give the similar result.
This approach is better than normal because it can test if a plane is tangent to a Quadric Surface, or if two quadric surfaces are tangent (but we can't know if internally or externally tangent). If there is only one common point between a plane and a quadric, their section $ax^2+bxy+cy^2+dx+ey+f=0$ (assume we eliminate z) should follow:
$$f=\frac{d^2}{4a}+\frac{e^2}{4c}$$
and $b=0$.
A: We are given the sphere
$$
(x+1)^2+(y-3)^2+z^2=25\tag1
$$
and the plane
$$
4x+3z+29=0\tag2
$$
The plane that is tangent at the point $p$ to a sphere centered at $c$ is
$$
(x-c)\cdot(p-c)=|p-c|^2\tag3
$$
$(1)$ says that
$$
c=(-1,3,0)\tag4
$$
$(2)$ can be written as
$$
x\cdot(4,0,3)=-29\tag5
$$
and combining $(4)$ and $(5)$ yields
$$
(x-c)\cdot(4,0,3)=-25\tag6
$$
we can scale $(6)$ to make the right side the square of the length of the vector on the right of the left side. In this case, we multiply by $-1$:
$$
(x-c)\cdot\overbrace{(-4,0,-3)}^{p-c}=\overbrace{\quad25\quad}^{|p-c|^2}\tag7
$$
Since $(7)$ says $|p-c|^2=25$, $p$ is on the sphere in $(1)$. Using $p-c=(-4,0,-3)$ and $(4)$, we get
$$
p=(-5,3,-3)\tag8
$$
