Four Orthogonal Circles on a Sphere

Question

Let $b$ and $c$ be two circles on a sphere, and $A$ be one of their intersections. We shall call $b$ and $c$ "orthogonal to each other" as the tangents of $b$ and $c$ at point $A$ are perpendicular to each other.

Let $a$, $b$, $c$ be three circles on a sphere. It is possible that each pair of circles from $a$, $b$, $c$ can be orthogonal to each other, and their intersections can form a triangle $\Delta ABC$ such that $\angle BAC = \angle CBA = \angle ABC =90^ {\circ}$:

So my questions are:

• Is it possible to create four circles $a$, $b$, $c$, $d$, so that each of two circles are pairwise orthogonal?

• If yes, can we create five circles with such characteristics? six, seven? Is there a maximum limit to their number ?

• If not, why not?

Answer 1

Your definition that circles are perpendicular if their tangents are perpendicular is appropriate.

The answer to your first question – "is it possible to create four circles on a sphere, $a$, $b$, $c$, $d$, so that each two circles are perpendicular to each other?" – is yes. Circles of latitude and longitude meet at right angles, for instance. Such "quadrilaterals" are never radially symmetric, though, so perhaps that's not quite satisfactory.

Imagine, then, a square pyramid intersecting your sphere, with its apex centered above what I shall call the north pole. The angles of the circles around the north pole made by its face planes are acute. As we move the pyramid downward, the angles increase, approaching the angles of the base, which are right angles. Unfortunately, the limit is reached precisely as the apex reaches the north pole, where the circular segments vanish. However, note that there is another circular polygon cut where the face planes reemerge from the sphere. The angles of this polygon are equal to the angles of the polygon about the north pole. So, when the apex reaches the north pole, these angles are all right angles.

What we have proved is that the face planes of square pyramid with its axis on a diameter of a sphere and its apex on the sphere cut four perpendicular circles in the sphere.

The answer to your second question – whether we can create such right-angled "polygons" with any number of sides – is also yes. We can also use pyramids to see this. Consider a right regular $n$-gonal pyramid for $n \geq 5$ intersecting the sphere like our square pyramid. If the apex is placed so that the lateral edges of the pyramid are tangent to the sphere, then the circles of intersection will make angles of measure zero with each other. As we move the pyramid down, the angles increase. As the apex approaches the north pole, the angles approach the interior angles of the polygonal base, which are obtuse. So, somewhere in between, the angles must be right angles.

That's an intuitive argument with some unproved assertions, but it isn't difficult to make it more precise, depending on the level of rigor you're wanting.

Note, incidentally, that the answer to all your questions is no if you're requiring the circles to be great circles, that is, circles whose centers are the center of the sphere. Such circles are the "lines" of spherical geometry.

Answer 2

Trying to understand your question in full. Pending your clarification in future comments the answer to first two questions is yes and for third question also, we can have infinitely many orthogonally intersecting continuous circle chains.

You mentioned a spherical triangle ( e.g., Equator with Greenwich and Bangladesh longitudes ), three vertices each $90^{\circ}$ with a spherical excess $90^{\circ}.$

Just like nice sketches provided by Costa in his answer, I also understand the circles on sphere should be continuous but not a spherical triangle with corner right angles.

Propose $n$ number of orthogonal circles ( can be called that way ) that must be all small circles required to be symmetrically arranged on a sphere symmetrically around (North-South) Polar axis.In this scheme the ratio of radius of eccentric circle to center offset is given by:

$$\frac{a}{h}=\sqrt{2} \sin \frac{\pi}{n} $$

The intersections are "diangles" enclosed within corner tangents making $90^{\circ}$ shown red.

The visualization/ drawing of such circles can be well realized by stereographic projection from plane containing polar patterns touching the South Pole projected on sphere by rays emanating from North Pole as shown. As a property of this projection, curves intersecting at $ 90^{\circ}$ in the ground plane patterns do so as well on the sphere.

Also in the light projection of (two sets) orthogonal small circles through North Pole we have a orthogonal shadow grid projected on South Pole Tangent plane. Is such a related formation perhaps what you are interested in?

EDIT1

The answers to your first two questions is Yes. ( third answer does not apply ).

I am including images of two different radius ratio polar rosettes in a square mode. Pentagonal/Hexagonal etc. rosettes for any $ n $ is possible.

EDIT2:

Also a five Circle rosette of orthogonal circles is shown here:

Geodesic curvature differential relation is used; small circle radius of curvature to great circle radius ratio is kept constant.