Newbie and systems of quadratic equations

Richard Owlett wrote:
> There are "better" ways to solve my problem.
> BUT, part of my goal is to refresh some math I haven't seen for almost 
> half a century.
> 
> I have a system of equations
> (x-a1)^2 +(y-b1)^2 +(z-c1)^2 = r1^2
> (x-a2)^2 +(y-b2)^2 +(z-c2)^2 = r2^2
> (x-a3)^2 +(y-b3)^2 +(z-c3)^2 = r3^2
> 
> which expand to
> x^2-2*a1*x+a1^2 +y^2-2*b1*y+b1^2 +z^2-2*c1*z+c1^2 = r1^2
> x^2-2*a2*x+a2^2 +y^2-2*b2*y+b2^2 +z^2-2*c2*z+c2^2 = r2^2
> x^2-2*a3*x+a3^2 +y^2-2*b3*y+b3^2 +z^2-2*c3*z+c3^2 = r3^2
> 
> [The a's, b's, c's, and r's for each problem are constant but will be 
> different from problem to problem, so I wish a symbolic solution.]
> 
> These can be rewritten as
> f1(x,y,z)=r1^2 -r2^2
> f2(x,y,z)=r2^2 -r3^2
> f3(x,y,z)=r3^2 -r1^2
> 
> which are convienently linear in x, y, and z.
> 
> Can Maxima do that subtraction for me?

You mean something like this:

(%i1) display2d:false;
(%o1) false
(%i2) (x-a1)^2+(y-b1)^2+(z-c1)^2=r1^2;
(%o2) (z-c1)^2+(y-b1)^2+(x-a1)^2 = r1^2
(%i3) (x-a2)^2+(y-b2)^2+(z-c2)^2=r2^2;
(%o3) (z-c2)^2+(y-b2)^2+(x-a2)^2 = r2^2
(%i4) (x-a3)^2+(y-b3)^2+(z-c3)^2=r3^2;
(%o4) (z-c3)^2+(y-b3)^2+(x-a3)^2 = r3^2
(%i5) expand(%o3)-expand(%o2);
(%o5) 
-2*c2*z+2*c1*z-2*b2*y+2*b1*y-2*a2*x+2*a1*x+c2^2-c1^2+b2^2-b1^2+a2^2-a1^2
         = r2^2-r1^2

Ray