In spherical geometry, are disks convex, and are chords internal to a circle?

Question

In Euclidean geometry, a chord of a circle is internal to the circle. However, the proof of this relies on the External Angle Theorem, which does not hold for spherical geometry.

In spherical geometry:

• (Spherical) Circles are intersections of planes with spheres
• (Spherical) Lines are great circles
• (Spherical) Chords are segments of great circles

Can a spherical chord exit a spherical circle?

If yes: Can you provide an example? If no: Can you prove this?

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Update

As requested:

Euclid proves in Proposition III.2 that "If two points are taken at random on the circumference of a circle, then the straight line joining the points [i.e. chord] falls within the circle." Visually, this seems obvious; but of course, it needs proof, which Euclid supplies.

Perhaps a more contemporary way of stating that is: All disks are convex, with a disk meaning a circle and its interior and the definition of convex being: a region is convex if for any two points in the region, the segment between those two points is a subset of the region.

So far, so good. What about non-Euclidean geometry? Visually, this claim, that disks are convex, would seem to apply to spherical geometry as well. I can think of no counterexample. But Euclid's proof fails in spherical geometry!

Specifically, his proof makes use of the exterior angle theorem, which does not apply to spherical geometry (see discussion here). So, we have the following:

1. Visually, disks in spherical geometry appear to be convex
2. All examples of spherical disks I can think of are convex
3. Yet the proof of convexity of the disk fails in spherical geometry

That would suggest that either:

1. There's an alternate proof of the convexity of the disk that does not require the External Angle Theorem
2. There are spherical disk that are indeed not convex

I request either a proof of #1 or an example of #2.

Answer 1

Here is a spherical disk that is non-convex:

(Image courtesy of Wikipedia)

How? you may ask. That disk is clearly convex! you might protest. To which I say: Sure, the small one inside the circle is, but I'm talking about the big one outside the circle!

In spherical geometry, the plane is bounded, so the distinction between "inside" and "outside" disappears, and it must be specified for every closed curve. Intuitively, we use the smaller part when discussing the nature of shapes on spheres, because that's how it works in unbounded spaces, but we don't have to in bounded ones. And if we aren't careful to specify that we are, some pedant can come along and say we aren't.

This also applies to chords, by the way. Whatever great circle (intersecting the small one, of course. Back, pedants!) you draw on the above sphere, focusing on the smaller part will make the small disk "convex" and the big disk "concave", while the bigger part of the line will do the opposite, and suddenly the small disk is the "concave" one!

Answer 2

It can happen topologically. On a plane, if you deform a circle into another circle and maintain the chord as a geodesic path between the endpoints, the chord remains internal to the circle. On a sphere, however, the deformation can cause the intended circle to pass through a great circle/line, so reversing orientation, and then the chord under this deformation is external to the reversed circle.