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I have a system of equations, see below. I wonder if it is possible to solve this for three unknowns $x,y, \gamma$ (the $c_i$ are known constants) in a computer program for ex. maple? I do not wanna spend time doing this by hand.

\begin{cases} \left(x - c_1 \right)^{2} + \left( y - c_2 \right)^{2} = (\gamma+c_3)^{2} \\ \left(x - c_4 \right)^{2} + \left( y - c_5 \right)^{2} = \left( \gamma + c_6 \right)^{2} \\ \left(x - c_7 \right)^{2} + \left( y - c_8 \right)^{2} = \left( \gamma + c_9 \right)^{2} \\ \left(x - c_{10} \right)^{2} + \left( y - c_{11} \right)^{2} = \left( \gamma + c_{12} \right)^{2} \end{cases}

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  • $\begingroup$ For almost all combinations of parameters, no solution will exist. $\endgroup$
    – Paul
    Feb 12 '16 at 13:07
  • $\begingroup$ I didnt specify variables who are where the unknown, have done it now. This comes from a problem in real life, so it will have one solution. $\endgroup$
    – Olba12
    Feb 12 '16 at 13:10
  • $\begingroup$ Write it as a matrix. $\endgroup$ Feb 12 '16 at 13:16
  • $\begingroup$ 3 unknowns, 4 independent equations...linear, or nonlinear, it's still unlikely to lead to a solution. $\endgroup$
    – Paul
    Feb 12 '16 at 13:18
  • $\begingroup$ @Olba12: For the benefit of your readers, kindly try to make your notation as simple as possible. (I made the edit.) However, as pointed out by others, you have $3$ unknowns but $4$ equations, So your $x,y,\gamma$ will be solvable only if your $c_i$ obey a constraint. In other words, one of the $c_i$ will have to serve as a fourth unknown and depend on the others. $\endgroup$ Feb 13 '16 at 4:48
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Here's one way to look at it. Given known $c_i$ and unknown $x,y,z$,

$$(x-c_1)^2+(y-c_2)^2 = (z-c_3)^2\tag1$$

$$(x-c_4)^2+(y-c_5)^2 = (z-c_6)^2\tag2$$

$$(x-c_7)^2+(y-c_8)^2 = (z-c_9)^2\tag3$$

$$(x-c_{10})^2+(y-c_{11})^2 = (z-c_{12})^2\tag4$$

Expand, then subtract $(1)$ from the others, and all second powers will be cancelled out. You'll get the simpler,

$$d_1x+d_2y+d_3z+d_4=0\\ d_5x+d_6y+d_7z+d_8=0\\ d_9x+d_{10}y+d_{11}z+d_{12}=0\tag5$$

where the $d_i$ are just expressions in the $c_i$. One can then solve for $x,y,z$.

However, substituting these into $(1),(2),(3),(4)$, you will find they will be satisfied only if the twelve $c_i$ obey a single constraint. In other words, one of the $c_i$ will depend on the others.

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