Number to string variable

Actually, they’re not strings but symbolic expressions.

This is how I would do it:

syms x y z r

x1=169;
x2=169;
y1=200;
y2=178;
z1=202;
z2=196;
x3=169;
x4=169;
y3=198;
y4=180;
z3=204;
z4=198;
equ1=subs((x-x1)^2+(y-y1)^2+(z-z1)^2-r^2);
equ2=subs((x-x2)^2+(y-y2)^2+(z-z2)^2-r^2);
equ3=subs((x-x3)^2+(y-y3)^2+(z-z3)^2-r^2);
equ4=subs((x-x4)^2+(y-y4)^2+(z-z3)^2-r^2);
[x,y,z,r]=solve([equ1,equ2,equ3,equ4], [x,y,z,r])

The substitutions are correct, but the problem is that the solutions are empty symbolic variables. There seems to be no analytic solution to the system.

You might want to give it a go with fsolve:

x1=169;
x2=169;
y1=200;
y2=178;
z1=202;
z2=196;
x3=169;
x4=169;
y3=198;
y4=180;
z3=204;
z4=198;
equ = @(b) [(b(1)-x1).^2+(b(2)-y1).^2+(b(3)-z1).^2-b(4).^2;
            (b(1)-x2).^2+(b(2)-y2).^2+(b(3)-z2).^2-b(4).^2;
            (b(1)-x3).^2+(b(2)-y3).^2+(b(3)-z3).^2-b(4).^2;
            (b(1)-x4).^2+(b(2)-y4).^2+(b(3)-z4).^2-b(4).^2];

B = fsolve(equ, randi([80 120], 4, 1))

produces (with one set of initial estimates for [x,y,z,r]):

B =

     303.8357
     199.3227
     135.3456
     150.2771

It stops after 400 iterations because it is still searching for a solution, so experiment with the options structure and other fsolve variations if necessary to give the best parameter estimates and information about the solution.

EDIT —

If you want to print the equations out as strings, use this (or something like it):

x1=169;
y1=200;
z1=202;
equstr ='(x-%.0f)^2+(y-%.0f)^2+(z-%.0f)^2-r^2';         % Produces String For ‘equ’
fprintf(1, [equstr '\n'], x1, y1, z1)

that produces:

(x-169)^2+(y-200)^2+(z-202)^2-r^2