Intersection of Three Spheres

Date: 06/05/2003 at 08:48:12
From: Rob
Subject: Intersection of three spheres

How do you find the intersection of three spheres?

(x-x1)^2 + (y-y1)^2 + (z-z1)^2 = r1^2
(x-x2)^2 + (y-y2)^2 + (z-z2)^2 = r2^2
(x-x3)^2 + (y-y3)^2 + (z-z3)^2 = r3^2

Point 1 (x1,y1,z1), point 2 (x2,y2,z2), and point 3 (x3,y3,z3) are 
known. I tried to solved them simultaneously using Cramer's rule but 
without any success. The equations get more complex.

Date: 06/05/2003 at 09:36:30
From: Doctor George
Subject: Re: Intersection of three spheres

Hi Rob,

Thanks for writing to Doctor Math.

I would recommend a vector approach to finding the intersection points 
of three spheres.

The basic strategy has two steps.
1. Find circle C that is the intersection of spheres 1 and 2.
2. Find the intersection of circle C with sphere 3.

The last section in the following answer from the Archives shows a 
vector approach to finding the intersection of two circles.

   Intersecting Circles
   http://mathforum.org/library/drmath/view/51710.html 

The method described here can be used to find the center point and 
radius of circle C. In step 2, look at the intersection of the plane 
of circle C and the third sphere. You should see that step 2 can be 
reduced to the intersection of two circles.

Does that make sense? Write again if you need more help.

- Doctor George, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 06/05/2003 at 09:58:57
From: Doctor Jerry
Subject: Re: Intersection of three spheres

Hi Rob,

The intersection of two spheres is a circle, and if you throw in a 
third sphere you'll often find two points as the intersection of all 
three.  

Cramer's Rule won't work because it applies to linear systems, not 
quadratic.

You can subtract equation 1 from eq 2, finding a linear relation in 
x,y, and z. Do the same for eq 1 and eq 3. Solve the linear system of 
two equations for x and y in terms of z. Substitute these results in 
any of the equations of the spheres. This will give an equation in 
terms of z alone. It will be quadratic. Solve it for z. Generate the 
corresponding x and y by going back to the linear system.

For x1 = y1 = z1 = 0; r1 = 3;  x2 = 1; y2 = 1; z2 = 1; r2 = 2; 
x3 = -1; y3 = 1; z3 = 2; r3 = 2

I found the linear system to be 

   -8 + 2 x + 2 y + 2 z = 0

  -11 - 2 x + 2 y + 4 z = 0

and solving this for x and y,

   x = (1/4)*(-3 + 2*z), y = (1/4)*(19 - 6*z)

Substituting into the first sphere equation, I found 

   z = (1/14)*(30 - Sqrt[109]),  z = (1/14)*(30 + Sqrt[109])

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/