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}
Best Answer
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.