<p>This looks like an instance of Robust Ellipse Fitting. Check this paper: Outlier Elimination for
Robust Ellipse and Ellipsoid Fitting <a href="http://arxiv.org/pdf/0910.4610.pdf" rel="nofollow">http://arxiv.org/pdf/0910.4610.pdf</a>.</p>

<p>A first rough and easy solution is provided by the ellipse of inertia (2D version of the ellipsoid of inertia <a href="http://en.wikipedia.org/wiki/Moment_of_inertia#Inertia_ellipsoid" rel="nofollow">http://en.wikipedia.org/wiki/Moment_of_inertia#Inertia_ellipsoid</a>). Its center is just the centroid and axes are given by Eigen vectors/values of the 2x2 matrix of inertia.</p>
center of a cluster of points and track shape
<p>I have plots of points which look like this.<img src="https://i.stack.imgur.com/UTAgq.jpg" alt="x-y plot1"> <img src="https://i.stack.imgur.com/TejgH.jpg" alt="x-y plot2">
<img src="https://i.stack.imgur.com/vPoxs.jpg" alt="enter image description here"></p>

<p>The tracks which these points form can be a circle or an ellipse. Clearly the center of the circular tracks in the two images above are different.</p>

<p><strong>How can I find the center point of these tracks (circular/elliptical)?</strong> I want to find the (x,y) coordinates which is the center, not necessary that it has to be a point that's in the plotted data set. i.e., I don't want a medoid.</p>

<p>EDIT: Also, <strong>is there anyway that I can find an equation for circle/ellipse that envelopes a majority of these points?</strong> In the elliptical track, I've added an ellipse that envelopes the points on the track. The values were calculated by trial and error. The center was also calculated by eye balling the plot. How can I do this programmatically?</p>