Intersection of two spheres

Author:Rafael Losada Liste

Topic:Algebra, Congruence, Constructions, Geometry, Intersection, Quadratic Equations, Sphere, Geometric Transformations

This activity belongs to the GeoGebra book Linkages. The degrees of freedom of a configuration can vary in particular cases. Let's look at a simple example. Let there be 4 points C, D, E and F that are one unit away from two points A and B. We will assume that point A remains fixed at the origin of coordinates, while point B can slide along the Z axis [to simplify, between (0,0,0) and (0,0,2)]:

C_x² + C_y² + C_z² = 1
D_x² + D_y² + D_z² = 1
E_x² + E_y² + E_z² = 1
F_x² + F_y² + F_z² = 1
C_x² + C_y² + (C_z - B_z)² = 1
D_x² + D_y² + (D_z - B_z)² = 1
E_x² + E_y² + (E_z - B_z)² = 1
F_x² + F_y² + (F_z - B_z)² = 1

We construct the spheres with centers A and B and radius 1. Their intersection will be, in general, a circle c, with a variable radius depending on the position of B. In it we place the four points, C, D, E and F. The configuration thus obtained has (remember that we assume A is fixed), in general, 5 degrees of freedom (necessary to determine the position of each point B, C, D, E and F). Note that the above system has 8 equations and 13 unknowns. But if we position B at two particular points, this can change. If we set B = (0,0,2), the degrees of freedom are reduced to 0, since the four points C, D, E and F are forced to occupy the position (0,0,1), the only (real) solution of the previous system of equations (because an equation of the type x² + y² = 0 only has the null solution, if we consider x and y as real numbers). The configuration becomes rigid (in the field of real numbers). [The same would happen at (0,0,-2).] On the other hand, if we fix B = (0,0,0) the position of B coincides with that of A, so points C, D, E and F are free to move on the sphere with center A and radius 1. In total, 8 degrees of freedom. Note that the last four equations of the previous system coincide with the first four, so they do not provide more information. In order to best solve the continuity problem of definition of points C, D, E and F, in the construction, every time we make B coincide with A, points C, D, E and F, move the position they had at c (which remains indeterminate as the intersection of two spheres) to the equator of the sphere centered at A. And vice versa, every time B leaves the position (0,0,0), each of those points C, D, E and F, up to that free instant on the sphere, projects its position on the circle c.

Authors of the construction of GeoGebra: Carlos Ueno and Rafael Losada

Intersection of two spheres

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