Intersection of sphere and plane is a point, find c, such that (0,0,c) is center of sphere

Question

Sphere: $$ x^2+y^2+(z-c)^2=1 $$ Plane: $$ x+2y+3z=0 $$ Find the values of $c$, for which the intersection of the sphere and the plane is a point.

Well, I know that the sphere has the center (0,0,c) and radius 1. And the plane has the point (0,0,0) and vector normal to the plane $(1,2,3)$.

But I'm stuck. I just don't know where to start.

Should I consider the plane tangent to the sphere? If I can do that, then I think I can find c.

Answer 1

Hint

Yes, the sphere and plane are touching tangentially. It is not a point but a zero radius circle.

Consider multiplicity of roots and vanishing discriminant.

EDIT 1

We can see how the projection behaves.

Eliminating x, the projection is represented by a function between y and z as:

$$(2 y + 3 z)^2 = y^2+(z-c)^2 $$

$$ 3 y^2 +8 z^2+12 yz +c(2z-c) =0 $$

This should be a degenerate case becoming a point or a pair of straight lines for a certain c , but right now cannot see how. It can be ellipse for c=0, the problem is uninteresting.

Answer 2

I'm not realy sure about my answer but if you move 1 unit from the center of the sphere in the direction (and opposite direction) of the normal vector you should reach a point on the plane. then just substitude the point on the equation of the plane. I got $ c=+14/3$ for the one in the direction of the normal vector