Under what conditions of $(h, k, l, r, a, b, c, d)$ will plane $ax + by + cz = d$ intersect the sphere $(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2$?
My instinct tells me to find the distance from the plane to the center point of the sphere and compare that to the radius of the sphere, but I do not remember my vector calc very well. Requires finding normal vector to plane, but I do not know what to do with that information. Thanks in advance
The distance from the center $(h,k,l) $ to the plane $ax+by+cz-d=0$ is
$$D=\frac {|ah+bk+cl-d|}{\sqrt {a^2+b^2+c^2}} $$
If $D>r $ no intersection. If $D=r $, the plane is tangent to the sphere.
If $D <r $, the intersection is a circle of radius $r'=\sqrt {r^2-D^2} $.