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Given three circles centered a points $A(\frac{1}{2},1)$, $B(1,0)$ and $C(0,0)$ with radii of $1$, as described in the following diagram:

enter image description here

The objective is to compute the 2D trilateration point. The point $P$ in the digram is a possible solution, based on it being the intersection point of the lines constructed from the intersection points between circle pairs.

Using the 'corrected' solution provided in the post, I setup the following set of equations:

$$ (P_x - A_x)^2 + (P_y - A_y)^2 = R_A^2\\ (P_x - B_x)^2 + (P_y - B_y)^2 = R_B^2\\ (P_x - C_x)^2 + (P_y - C_y)^2 = R_C^2\\ $$

Which then further simplifies to the following two equations: $$ \begin{align*} 2P_x + 0P_y = 1 \\ -1P_x + 2P_y = \frac{1}{4}\\ \end{align*} $$

which can also be expressed as: $$ \left[\begin{matrix} P_x\\ P_y \end{matrix}\right] \left[\begin{matrix} 2 & 0\\ -1 & 2 \end{matrix}\right] = \left[\begin{matrix} 1\\ \frac{1}{4} \end{matrix}\right] $$

Rearranging to solve for $P_x$ and $P_y$:

$$ \left[\begin{matrix} P_x\\ P_y \end{matrix}\right] = \frac{1}{4} \left[\begin{matrix} 2 & 0\\ 1 & 2 \end{matrix}\right] = \left[\begin{matrix} \frac{1}{2}\\ \frac{3}{8} \end{matrix}\right] $$

What I'm trying to get my head around, is that the initial set of equations was laid to to determine the values of $P_x$ and $P_y$ based on the constraint that that they hold true as being of length $R_i$ away from the given centers $A$, $B$ and $C$. $P$ is not $1$ unit away from any of the centers but rather $\frac{5}{8}$ units. The determinant is non-zero and as far as I can tell the matrix is not rank deficient.

If one were to plug in the values obtained back into the first equation the results don't add up:

$$ (P_x - A_x)^2 + (P_y - A_y)^2 = R_A^2\\ (\frac{1}{2} - \frac{1}{2})^2 + (\frac{3}{8} - 1)^2 \neq 1\\ $$

My question is did the transposing (subtracting one set of equations from the other) cause the final result not to be correct?

I was hoping that for situations described in the above diagram, either the determinant would be zero or the relating matrix would be rank deficient, or some other kind of easily verifiable indicator


The result is obviously the intersection between lines constructed from intersections, but what can be said about the nature of $P$ - is it optimal in some respect? I've plotted the Torricelli point as $T$, which minimizes the sum distance between the centers and itself - $P$'s sum is actually $0.54\%$ longer.

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  • $\begingroup$ Could someone with rep please embed the image, as I don't have enough points as of yet. $\endgroup$ – Sari T. Mar 3 '17 at 23:46
  • $\begingroup$ The post you link gives the solution to the intersection of $3$ circles which have a common point. Your circles don't have a common point, so those equations don't apply. $\endgroup$ – dxiv Mar 4 '17 at 0:11
  • $\begingroup$ @dxiv Sure I understand that, but wouldn't there be some indication, such as a problematic determinant or rank issue with the matrix or some such, that one could use to realize the solution wont resolve? $\endgroup$ – Sari T. Mar 4 '17 at 0:24
  • $\begingroup$ That's very clear now. thank-you! it'd be great if you post that as an answer. $\endgroup$ – Sari T. Mar 4 '17 at 1:08
  • $\begingroup$ Glad it helped. I converted my previous comment to an answer. $\endgroup$ – dxiv Mar 4 '17 at 1:25
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Using the 'corrected' solution provided in the post, I setup the following set of equations

The post in question gives the solution to the intersection of $3$ circles which have a common point. Your circles don't have a common point, so those equations don't apply. (And in fact, as you noted, the system as written has no solutions.)

I was hoping that for situations described in the above diagram, either the determinant would be zero or the relating matrix would be rank deficient, or some other kind of easily verifiable indicator

There is no such indication in general, nor is one expected to (necessarily) exist. After subtracting the equations, the resulting linear system is no longer equivalent to the original system. Moreover, the linear system does not (and could not) have any "memory" of what the original equations were, and whether they were consistent or not.

For a simplistic example, consider the system: $$ \begin{cases} \begin{align} x^2 +y^2 &= 0 \\ x^2 +(y−1)^2 &=0 \end{align} \end{cases} $$

Obviously enough, the system has no solutions, yet if you subtract the equations you get the "full rank" linear equation $2y−1=0\,$. Only, when you attempt to substitute that "solution" back into the original equations, it doesn't work out.

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