Find all planes which are tangent to a surface I'm given the surface $z=1-x^2-y^2$ and must find all planes tangent to the surface and contain the line passing through the points $(1, 0, 2)$ and $(0, 2, 2).$ I know how to calculate tangent planes to the surface given one point, but how would I do so given two points? Any push in the right direction would be appreciated. Thank you.
 A: tangent plane at $(a, b, 1 - a^2 - b^2)$ is 
$$2a(x-a)+2b(y-b) + (z +a^2 + b^2 - 1) = 0  $$
since the point $(1,0,2)$ is to be on the plane
we get $2a(1-a) + 2b(-b) + (a^2 + b^2 + 1) = 0$ which can be simplified to yield
$a^2 + b^2 -2a - 1 = 0.$ 
 in the same way making the point $(0,2,2)$ on the plane requires
$2a(-a) + 2b(2-b) + (1+a62 + b^2) = 0.$ on using the previous constraint 
we have $$a = 2b, 5b^2-4b - 1 = 0 $$ 
so there are only two points $(2, 1, -4), (-2/5, -1/5, 4/5)$ so that the tangent plane contains the line through  $(1,0,2)$ and $(0,2,2)$
A: Compute the plane by the three points $(x,y,z)$, $(1,0,2)$, $(0,2,2)$ and check that it is tangent to the surface at $(x,y,z)$.
The normal vector to the plane is given by $N_p=(x-1,y-0,z-2)\times(1-0,0-2,2-2)$.
The normal vector to the surface is given by $N_s=(2x,2y,1)$.
These two vectors are parallel, $N_p\times N_s=0$.
$$-x^2+4xy+y^2-4y-1=0,\\
2+2y^2+4x-2xy-2x^2=0,\\
-y(4x^2+4y^2+4)+2x(x^2+y^2+1)=0.$$
Cancelling the third component gives $x=2y$, then the first reduces to $(y-1)(5y+1)=0$, and the second to $-2(y-1)(5y+1)$.
Solutions: $(2,1,-4)$ and $(-\frac25,-\frac15,\frac45)$.
A: Given $p_1,p_2$ and $G(p) = 0$ with $p = (x,y,z)$, determine a plane $\Pi\to (p-p_1)\cdot\vec n=0$ with $\{p_1,p_2\}\in \Pi$ and such that $\Pi$ is tangent to $G$ at $p^*$
Representing for $\nabla G(p^*)$ the normal to the the surface $G(p)=0$ at point $p^*$ we have the conditions.
$$
\cases{
\nabla G(p^*) = \lambda\vec n\\
(p_2-p_1)\cdot \vec n = 0\\
G(p^*) = 0\\
(p^*-p_1)\cdot\vec n = 0\\
\|\vec n\| = 1
}
$$
seven equations and seven $(p^*,\vec n,\lambda)$ unknowns.
