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I want to find the solution space of coordinates for point $p$ that satisfies the following system: $$ \begin{cases} [distance(p,a) - distance(p,b)] = k_1\\ [distance(p,c) - distance(p,d)] = k_2 \end{cases} $$

where the points $a,b,c,d,q$ have arbitrary known positions, $k_1 = k - [distance(q,a) - distance(q,b)]$ and $k_2 = k - [distance(q,c) - distance(q,d)]$

I'm trying to solve it on 3D space, but for simplicity I started to analyse it on 2D to further extend my solution. I'm using Cartesian coordinates and $distance(a,b)$ represents the Euclidean distance between points $a$ and $b$.

1) Does system with this shape have a specific name?

2) How can I find an exact solution for this problem?

This is a non-linear system, and from what I searched usually non-linear systems can be solved by either linearization or numeric methods. I'm interested in an exact solution, so I would like to avoid numeric methods. I read that is not always possible to linearize systems of equation, but I couldn't find which are the conditions to know if it is possible or not.

3) How do I know if it is possible to linearize this system?

4) Would make the problem easier to solve if we add two additional arbitrary points $e,f$ to generate a third equation like $[distance(p,e) - distance(p,f)] = k_3, \mbox{ where } k_3 = k - [distance(q,e) - distance(q,f)]$?

Hyperbolas can be defined as the locus of points where the absolute value of the difference of the distances to the two foci is a constant equal to 2a, the distance between its two vertices. In this case I could see the problem as the intersection of hyperbolas $H_1$ and $H_2$ with foci $a,b$ and $c,d$, respectively.

$$ H_1 := [distance(p,a) - distance(p,b)] = k_1\\ H_2 :=[distance(p,c) - distance(p,d)] = k_2 $$

5) Is there any method to find intersections of conics section?

6) Would change the coordinate system make it easier to solve (ex. polar coordinates)?

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  • $\begingroup$ Point p is variable and (a,b,c,d,q) are fixed, right? I think q should not be included here. $\endgroup$ – Narasimham May 21 '16 at 15:15
  • $\begingroup$ Yes, they are fixed. I included q, because it affects both equations and it might give some dependency or relation between them. $\endgroup$ – cpnogueira May 21 '16 at 15:27
  • $\begingroup$ If so, is not $ distance(p,a) - distance(p,b)=k_1 ; \quad distance(p,c) - distance(p,d) =k_2 $ adequate? because $q$ is fixed? $\endgroup$ – Narasimham May 21 '16 at 16:03
  • $\begingroup$ Yes, it is adequade, but $k_2$ and $k_2$ depend directly in the position of $a,b,c,d$ and $q$. I edited the question to express in terms of $k_1$ and $k_2$, but I don't want to omit this relation, since $k_2$ can be rewritten as $k_2= k_1 + [distance(q,a)−distance(q,b)] − [distance(q,c)−distance(q,d)]$. I'm not sure how relevant this relation is. Do you think it is not relevant to solve the problem? $\endgroup$ – cpnogueira May 21 '16 at 16:52

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