There will generally be no common point of intersection between all 4 spheres, so where all the spheres intersect. Of course, you may get lucky. And, yes, you can just throw a tool like solve at it. Sometimes solve will work, or not. It won't really tell you why it failed, or how it solved the problem. It looks very much like you want to understand why and how to solve the problem, and I think that is what your question really comes down to, one of understanding.
%Spheres
S1 = [1;1;1];
S2 = [2;1;1];
S3 = [1.5; 0;1];
S4 = [1.5; 2;1];
r = 1;
% Sphere equations
syms x y z % These are the unknowns
EQ(1) = (x-S1(1))^2 + (y-S1(2))^2 + (z-S1(3))^2 == r^2;
EQ(2) = (x-S2(1))^2 + (y-S2(2))^2 + (z-S2(3))^2 == r^2;
EQ(3) = (x-S3(1))^2 + (y-S3(2))^2 + (z-S3(3))^2 == r^2;
EQ(4) = (x-S4(1))^2 + (y-S4(2))^2 + (z-S4(3))^2 == r^2;
The common trick is to subtract a pair of equations. What does that do? You will see in general that the squared terms in all of the variables disappear. And what does that mean? Imagine two spheres in some configuration where the two spheres intersect. Always, unless one sphere is inside the other, or they touch at a single point, the result will be a nice perfect circle, that lies in some plane.
TRY IT!
EQ12 = simplify(expand(EQ(1) - EQ(2)))
So we learn that spheres 1 and 2 intersect in the plane x = 3/2, with y and z extending parallel to the y and z axes. I'll draw those two spheres, so we can see what happens.
[x1,y1,z1] = sphere(1000);
H1 = surf(x1+S1(1),y1+S1(2),z1+S1(3));
H1.EdgeColor = 'none';
H1.FaceColor = 'r';
H1.FaceAlpha = 0.3;
hold on
% each sphere has the same unit radius with a different center,
% so I can just reuse x1,y1,z1
H2 = surf(x1+S2(1),y1+S2(2),z1+S2(3));
H2.EdgeColor = 'none';
H2.FaceColor = 'g';
H2.FaceAlpha = 0.3;
hold off
axis equal
So those first two spheres intersect along a plane, with x==3/2. As I said, the intersection is a circle, and that circle lies parallel to the y and z axes.
Do you see how this works? If not, go back and think about what I did before you go any further.
Next, we can look at other pairs of spheres. How do they interact?
EQ13 = simplify(expand(EQ(1) - EQ(3)))
Similarly, EQ13 defines the plane of the circle where the corresponding spheres intersect. This plane lies at an angle to the axes, but that is not important.
EQ14 = simplify(expand(EQ(1) - EQ(4)))
EQ23 = simplify(expand(EQ(2) - EQ(3)))
HOWEVER, these last two resulting equations imply a problem, as those two planes are parallel, but not coincident. They never intersect. And that proves there can be no common point of intersection for all 4 spheres.
As I originally suggested, that will usually be the case. At best, you might decide to find the point that best satisfies all of the equations. But there is NO common solution in this case. We just proved that. How might we automate this process? A simple approach might be to reduce all pairs of sphere intersections into linear equations.
k = 0;
for i = 1:4
for j = i+1:4
k = k + 1;
eqpairs(k) = simplify(expand(EQ(i) - EQ(j)));
end
end
[A,B] = equationsToMatrix(eqpairs,[x,y,z])
We now have a system of 6 linear equations, in three unknowns, of the form
A*[x;y;z]==B
Any solution that exists must satyisfy ALL 6 of those linear equations. Does an exact solution exist? That will be true ONLY is the vector B can be wrtten as a linear combination of the columns of A. We can learn that from rank.
rank(A)
rank([A,B])
So the matrix A has rank 2. But when as append the vector B to A, and again test the rank, the result has rank 3. Again, this proves no exact solution can exist.
Finally, I'll plot all 4 spheres, viewing from the top down, so we can visualize what happened, and why no solution existed.
H1 = surf(x1+S1(1),y1+S1(2),z1+S1(3));
H1.EdgeColor = 'none';
H1.FaceColor = 'r';
H1.FaceAlpha = 0.3;
hold on
H2 = surf(x1+S2(1),y1+S2(2),z1+S2(3));
H2.EdgeColor = 'none';
H2.FaceColor = 'g';
H2.FaceAlpha = 0.3;
H3 = surf(x1+S3(1),y1+S3(2),z1+S3(3));
H3.EdgeColor = 'none';
H3.FaceColor = 'b';
H3.FaceAlpha = 0.3;
H4 = surf(x1+S4(1),y1+S4(2),z1+S4(3));
H4.EdgeColor = 'none';
H4.FaceColor = 'y';
H4.FaceAlpha = 0.3;
hold off
axis equal
view(-90,90)
Do you see the yellow and blue spheres intersect at one point? They touch, just a brief kiss. But the red and green spheres intersect at a circle, that completely surrounds the intersection point of the yellow and blue spheres. So graphically, you can visualize that no solution exists.
Sorry.
