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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 

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 

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 

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

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 
Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

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