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}

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.