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 $2x+2y-z=0$

Question

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?

Answer 1

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 $z$.

Answer 2

There are several methods. I'll outline two here

1. 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.

2. 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 $(2,2,-1)$.