Newbie and systems of quadratic equations

Raymond Toy (RT/EUS) wrote:

> 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
> 
> 

YES! Thank you. That gives me the "brute force" tool which I wanted.

I neglected to ask if there was a reference for using Maxima to 
"symbolically" solve systems of linear equations. ( ie the a's and b's 
are constants but don't have a explicit numeric value)

I browsed some of the tutorials but didn't that seemed to be directed at 
whatever audience I'm typical of.