Pole of inaccessibility

Minimum Bounding Circle

gallery

On a sphere the circumcenter is also the intersection of the three perpendicular bisectors of the sides. The same proof from the planar case also works on the sphere. Namely, the perpendicular bisector is the locus of points equidistant from two given points. To find a point equidistant to 3 given points, take the loci of equidistant points from pairs of the given points, and then intersect them.

Jacob Rus

Sure, makes sense. My (admittedly ad-hoc) computation with Turf is giving three separate points of intersection. I think where it goes wrong is computing the bearing from A to B and adding 90° at the midpoint.

You can try this one: https://observablehq.com/d/f3f94ffbfbdc0185

Thanks, I was able to implement the compass and straightedge construction via Turf, so it works well enough for me. Maybe I'll study the spherical coordinate cross product later.