In my last post I described how it is possible to find the centre and radius of a sphere given four or more points on its surface and not belonging to the same plane.
What happens if the points do belong to the same plane?
Let me assume that the points belong to a circumference on the sphere (the intersection of the sphere with the plane). Be the points , with . The linear system
in the unknowns (see my previous post) is underdetermined (rank 3) and does not individuate one single sphere, but rather a pencil of spheres sharing the common circumference.
So in this conditions the method is not useful for finding a sphere, but it can be used to find the circumference in 3D containing the points: here is how.
The m x 4 matrix of the system, , can be decomposed in singular values as
where (m x 4) is the matrix having the left singular vectors as columns, (diagonal 4 x 4) is the matrix of the singular values and (4 x 4) is the matrix having the right singular vectors as columns.
Being 3 the rank of the matrix, one of the singular values is zero (within machine precision) and the corresponding right singular vector spans the nullspace of A.
As all the points lie on the circumference, the second hand vector lies in the range of A and a solution can be found as
remembering that and ; is the diagonal matrix having if and if .
As always with underdetermined system, there is an infinity of solutions, obtained by adding to (2) any vector in the nullspace. In this case, the nullspace is spanned by the right singular vector corresponding to the null singular value, say .
So is the parametric equation of the pencil of spheres sharing the circumference one is looking for.
Now, the circumference will be an equator of the smallest sphere in the pencil. So, remembering from my previous post that , , and ,
By equating to zero the derivative of the expression to be minimized one gets:
whence the coordinates of the centre and the radius of the circumference can be obtained, while the the plane of the circumference is orthogonal to .