# Circle of a sphere

A **circle of a sphere** is a circle that lies on a sphere. Such a circle can be formed as the intersection of a sphere and a plane, or of two spheres. A circle on a sphere whose plane passes through the center of the sphere is called a great circle; otherwise it is a **small circle**. Circles of a sphere have radius less than or equal to the sphere radius, with equality when the circle is a great circle.

## Contents

## On the earth[edit]

In the geographic coordinate system on a globe, the parallels of latitude are such circles, with the Equator the only great circle. By contrast, all meridians of longitude, paired with their opposite meridian in the other hemisphere, form great circles.

## Related terminology[edit]

The diameter of the sphere which passes through the center of the circle is called its **axis** and the endpoints of this diameter are called its **poles**. A **circle of a sphere** can also be defined as the set of points at a given angular distance from a given pole.

## Sphere-plane intersection[edit]

When the intersection of a sphere and a plane is not empty or a single point, it is a circle. This can be seen as follows:

Let *S* be a sphere with center *O*, *P* a plane which intersects *S*. Draw *OE* perpendicular to *P* and meeting *P* at *E*. Let *A* and *B* be any two different points in the intersection. Then *AOE* and *BOE* are right triangles with a common side, *OE*, and hypotenuses *AO* and *BO* equal. Therefore, the remaining sides *AE* and *BE* are equal. This proves that all points in the intersection are the same distance from the point *E* in the plane *P*, in other words all points in the intersection lie on a circle *C* with center *E*.^{[1]} This proves that the intersection of *P* and *S* is contained in *C*. Note that *OE* is the axis of the circle.

Now consider a point *D* of the circle *C*. Since *C* lies in *P*, so does *D*. On the other hand, the triangles *AOE* and *DOE* are right triangles with a common side, *OE*, and legs *EA* and *ED* equal. Therefore, the hypotenuses *AO* and *DO* are equal, and equal to the radius of *S*, so that *D* lies in *S*. This proves that *C* is contained in the intersection of *P* and *S*.

As a corollary, on a sphere there is exactly one circle that can be drawn through three given points.^{[2]}

The proof can be extended to show that the points on a circle are all a common angular distance from one of its poles.^{[3]}

## Sphere-sphere intersection[edit]

To show that a non-trivial intersection of two spheres is a circle, assume (without loss of generality) that one sphere (with radius ) is centered at the origin. Points on this sphere satisfy

Also without loss of generality, assume that the second sphere, with radius , is centered at a point on the positive x-axis, at distance from the origin. Its points satisfy

The intersection of the spheres is the set of points satisfying both equations. Subtracting the equations gives

In the singular case , the spheres are concentric. There are two possibilities: if , the spheres coincide, and the intersection is the entire sphere; if , the spheres are disjoint and the intersection is empty.
When *a* is nonzero, the intersection lies in a vertical plane with this x-coordinate, which may intersect both of the spheres, be tangent to both spheres, or external to both spheres.
The result follows from the previous proof for sphere-plane intersections.

## See also[edit]

## References[edit]

- Hobbs, C.A. (1921).
*Solid Geometry*. G.H. Kent. pp. 397 ff.

## Further reading[edit]

- Sykes, M.; Comstock, C.E. (1922).
*Solid Geometry*. Rand McNally. pp. 81 ff.