Find the equation of the two tangent planes to the sphere $x^2+y^2+z^2-2y-6z+5=0$ which are parallel to the plane $2x+2y-z=0$ Find the equation of the two tangent planes to the sphere $x^2+y^2+z^2-2y-6z+5=0$ which are parallel to the plane $2x+2y-z=0$
My Attempt
We need to find a point which is shortest distance from the plane to the sphere. Let it be $(a,b,c).$ Then Equation of the tangent plane at $(a,b,c)$ is given by the formula
$$2a(x-a)+2(b-1)(y-b)+2(c-3)(z-c)=0$$
Finding the diametrically opposite point of $(a,b,c)$. I can find the equation of another tangent parallel to the plane.
I don't know how to find the shortest distance from the plane to the sphere.
Is there any short method to find the tangents?
 A: There are several methods. I'll outline two here

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*Find the center of the sphere. Take the line in the direction $(2,2,-1)$ (the normal vector of the plane) that goes through the center. It will intersect the sphere in the two points you are after.


*Given any surface defined by an equation $f(x,y,z)=0$ (where $f$ is suitably differentiable), and any point on the surface, the gradient of $f$ at that point is a normal vector to the surface at that point. We want places on the sphere where the normal vector is parallel to $(2,2,-1)$ (note that this is the gradient of the LHS of the plane equation, same deal there). Which is to say, we want places where the gradient of $x^2+y^2+z^2-2y-6z+5$ is parallel to $(2,2,-1)$.
A: Instead complete the square
$$x^2+(y-1)^2+(z-3)^2 = 5$$
then take the gradient
$$\langle x, y-1, z-3 \rangle = \lambda\langle 2, 2, -1\rangle \implies x = y-1 = \frac{3-z}{2}$$
which means
$$x^2 + x^2 + 4x^2 = 6x^2 = 5 \implies x = \pm \sqrt{\frac{5}{6}}$$
This gives you your two points once you plug in for $y$ and $z$.
