Error of minimum distance to intersection of circles I am trying to find a point in 2D space via a type of triangulation measurement. Each measurement gives me a circumference on which my point could be. After several different measurements, I have several circumferences which would, ideally, intersect at my point location.
To find the most likely location, I have numerically calculated the sum of distances from all points in my 2D grid to every circle, which returns some form of distribution of distance sums.
Is there some 'most appropriate' way that I can turn these distance sums into probabilities of location of the intersection point?
Or, alternatively, is there a more appropriate way to fit the location of my point to the intersection of the circumferences I have measured? 
 A: For an unknown point $p_0=(x_0,y_0)$ in the plane, several circles with center $p_1...p_n$ are given, with the following properties


*

*$p_0$ is equal to $p_i+r_ie_i(\theta_i)$,

*$e_i(\theta_i)=[\cos(\theta_i) \sin(\theta_i)]$ is a unitary vector in $R^2$, with random angle $\theta$ with unknown distribution and media zero,

*$r>0$ a random variable, with media equal to $r_0$ known, and distribution unknown (perhaps with low deviation?). 

*$r$ and $\theta$ are assumed to be independent.


Assuming the distribution of $\theta$ has media zero, then our best estimator is:
$$
\hat p_0=\frac 1n \sum_{i=1}^n p_i+r_ie_i(\theta_i)\\
=\frac 1n \sum_{i=1}^n p_i+\frac 1n \sum_{i=1}^nr_ie_i(\theta_i)\\
=\frac 1n \sum_{i=1}^n p_i+\frac 1n \sum_{i=1}^nr_i[\cos(\theta) \sin(\theta)]\\
=\frac 1n \sum_{i=1}^n p_i\\
$$
If we cannot assume the angle has media zero, the problem turns much more complicated. 
Under that case, if we assume the following distribution properties for $r$ and $\theta$:


*

*$r_i=r_0+e_{r_i}$, $e_{r_i}$ with mean zero and unknown distribution

*$\theta_i=\theta_0+e_{\theta_i}$, $e_{\theta_i}$ with mean zero and unknown distribution


Then 
$$
x_0=x_i+(r_0+e_{r_i})\cos(\theta_0+e_{\theta_i})\\
=x_i+r_0 \cos(\theta_0+e_{\theta_i})+e_{r_i} \cos(\theta_0+e_{\theta_i})\\
y_0=y_i+(r_0+e_{r_i})\sin(\theta_0+e_{\theta_i})\\
=y_i+r_0 \sin(\theta_0+e_{\theta_i})+e_{r_i} \sin(\theta_0+e_{\theta_i})\\
$$
This do not have a closed solution for an estimator. 
AS alternative, you can setup the optimization problem for $x_0,y_0,\theta_0, e_{r_i}, e_{\theta_i}$:
$$
\mathcal{P}: \min \sum_{i=1}^n e_{r_i}^2+e_{\theta_i}^2\\
x_0=x_i+(r_0+e_{r_i})\cos(\theta_0+e_{\theta_i})\\
y_0=y_i+(r_0+e_{r_i})\sin(\theta_0+e_{\theta_i})
$$
A hard, but the best way for getting the estimator of $x_0,y_0$.
A: A typical approach is to try to find the point that minimizes the sum of the squares of the errors.  For any point in the grid you can compute the distance to each of your measurement points and the difference between the measured distance and the physical distance.  Square the errors and add them up.  Take the location that minimizes that sum of squares.  
Rather than calculating this sum for each point in the grid, for any pair of measurement points there will be two positions that would have zero error, the intersection of the circles drawn about each point with radius the measured distance.  Hopefully the correct points will all be clustered.  You can take three measurement points that are in a nice triangle and you should find one point that fits the three measurements reasonably closely.  You can then use a minimizer to find the best fit.
