Maximize length of side of triangle from points on a circle
You could transform the question into finding the biggest minimum angle of the triangle vertices relative to the centre.
Preparatory work:

Find centre $(x_C, y_C)$, if not already known (intersect bisectors of randomly-chosen pairs of points). $o(1)$

a) Make list of angle position for each point, $a_i = \text{atan2}(y_i-y_C, x_i-x_C)$. $o(n)$
b) Sort angle list. $o(n \log n)$


Main determination:

In succession for each point:
a)  find points closest to $\pm 120^\circ$ points  from the list, $o(\log n)$ per point
b) record max minimum angle with this point, update overall maximum if needed. $o(1)$ per point

Post process

Convert angle back into distance (either directly or by retrieving the indices of the appropriate points). $o(1)$

Explaining #3 a little more: you should find four points from the two binary searches, one either side of the $+120^\circ$ &amp; $-120^\circ$ points. That gives you four triangles to examine for angles (differences), composed of the base point plus choices from left/right of each of the $\pm 120^\circ$ lines.  (If any of the points are the same (few/crowded points) , there will be fewer triangles to check).

For convenience you might add extra copies of the points either side of the wraparound $0^\circ=360^\circ$ to avoid special cases in the search - eg. duplicate the point at (say) $+358^\circ$ to also be seen at $-2^\circ$.