How to define small circle on an ellipsoid

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I am trying to figure out the proper definition of a small circle on a biaxial ellipsoid of revolution. One definition is the intersection of the ellipsoid with a cone emanating from the center of the ellipsoid.

The other way I can imagine to define it is a plane intersecting the ellipsoid in which the plane does not also intersect the center of the ellipsoid (or else it would be a great circle).

This Wikipedia article also discusses sphere-intersection, but it is limited to spheres, and I am interested in ellipsoids.

Does anyone know if these two methods, cone intersection and plane intersection, result in the same curve? If not, which one is a small circle, and what would be the name of the other resulting curve?

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The only small circles that you can get are found by intersection with a plane orthogonal to the revolution axis, which are also the intersections with a coaxial cone with apex at the center. This can be sketched in 2D:

Intersection with other planes yield ellipses, and intersection with non-coaxial cones are non-planar curves, with no name.


Note that you can reason from a spherical model that you stretch in one direction. The small circles are formed by the intersections with planes or with cones of revolution. After stretching, the planes remain planes, but the cones become elliptic, unless their axis is in the direction of the stretching.

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The intersection with a plane is, in general, an ellipse. The intersection with a cone is a more complicated curve, in general not lying on a plane. If what you want is a planar circle all lying on the ellipsoid, then I'm afraid that is not possible, unless particular conditions are met (e.g. a rotational symmetric ellipsoid, intersected by a plane perpendicular to the axis of symmetry).

EDIT.

If what you want is a closed line whose points have all the same distance $r$ from a given point $P$ on the surface of the ellipsoid, you can obtain it from the intersection of the ellipsoid with a sphere of radius $r$ centred at $P$.