How to measure variance of distances from origin

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I'm trying to measure the sample variance of some data. Such data are 2D euclidean distances from the origin (0,0).

Supposing to have the 2 components X and Y used to calculate the distance, it's trivial to calculate variances $\sigma^2_x$ and $\sigma^2_y$. However the measurement system I'm evaluating gives me only a single scalar measurement that is the euclidean distance (2D) from the origin. These measurements are of course always positive and does not follows a gaussian distribution.

I'm looking for a single scalar value to measure dispersion around the origin, that maps something similar to an average of the variances $\sigma^2_x$ and $\sigma^2_y$, but (reasonably) mathematically correct to obtain some repeatability score of the measurement system (e.g. Cm or Cmk).

There are some conditions that allows us to consider both the components X and Y to have zero-mean and following a gaussian distribution with respectively variances $\sigma^2_x$ and $\sigma^2_y$. I can also make the assumption that the 2 components are independent, so the resulting 2D distribution can be seen as a bivariate centered Gaussian distribution with variances $\sigma^2_x$ and $\sigma^2_y$.

What can be the right way to calculate a measure of "repeatability" of these measurements, given some samples of the single scalar output "2D euclidean distance from (0,0)"?

I also found some interesting starting points here:

https://en.wikipedia.org/wiki/Noncentral_chi-squared_distribution

http://ballistipedia.com/index.php?title=Circular_Error_Probable

http://ballistipedia.com/index.php?title=Closed_Form_Precision

but I'm not able to come up with a solution to my problem. Can you help me?

Thank you in advance

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The answer to the title is straight forward. If you have some data in $R^2$ from which you calculate the Euclidean distance from the origin then you create some scalar vector which contains all those distances call this vector $d$, then if you want the variance of those distances you just have to calculate in the usual fashion $\sigma_d^2$. I do not get your confusion? (maybe I misinterpret something).