Given 3 spheres, find the equation of the plane that touches each of the spheres on the same side..?

Question

I have a problem I am trying to solve, but I have no idea how to solve it.

If I have 3 spheres, $A(1, 2, 0), B(4, 5, 0), \text{ and } C(1, 3, 2)$ of radius 1, how would I go about finding the equation of the form $ax + by +cz = d$ of the plane which touches each of the spheres on the same side..?

I have been racking my brain all day, but I can't seem to figure it out.

If someone wouldnt mind lending a hand, It would be much appreciated..

Thanks Corey

Answer 1

If they all have the same radius, the problem can be easily solved. The plane needs to be parallel to the plane that contains the three centers; This yields the parameters $a, b, c$. Choosing the right distance to the plane that contains the centers gives you the two possible values for $d$.

Answer 2

It is easy to find a formula for the perpendicular distance between a point and a plane. The formula will give negative and positive distances for points on opposite sides of the plane, which will deal with "the same side" part of the question, assuming you know what "the same side" means here. (I don't know what it means from your statement of the problem, but I could make a plausible guess).

So in the general case, the distance of the plane from the center of each sphere gives 3 equations. Your equation of the plane apparently has four unknowns, but the plane $kax + kby + kcz = kd$ is the same as the plane $ax + by + cz = d$ for any non-zero value of $k$.