# Circle of a sphere

(Redirected from Plane–sphere intersection)
Small circle of a sphere.
$BC^2=AB^2+AC^2$, where C is the center of the sphere, A is the center of the small circle, and B is a point in the boundary of the small circle. Therefore, knowing the radius of the sphere, and the distance from the plane of the small circle to C, the radius of the small circle can be determined using the Pythagorean theorem.

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.

## On the earth

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

The diameter of the sphere which passes though 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

When the intersection of a sphere and a plane is not empty or a single point, it is a circle.

That the intersection of a sphere and a plane is, in fact, a circle can be seen as follows. Let the 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 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 with center E.[1] Note that OE is the axis of the circle.

As a corollary, on a sphere there is exactly one circle that can be drawn though 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

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

$x^2 + y^2 + z^2 = R^2.$

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

$(x-a)^2 + y^2 + z^2 = r^2.$

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

\begin{align} (x-a)^2 - x^2 & = r^2 - R^2 \\ a^2 - 2ax & = r^2 - R^2 \\ x & = \frac{a^2 + R^2 - r^2}{2a}. \end{align}

In the singular case $a = 0$, the spheres are concentric. There are two possibilities: if $R = r$, the spheres coincide, and the intersection is the entire sphere; if $R \not= r$, 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.

## References

1. ^ Proof follows Hobbs, Prop. 304
2. ^ Hobbs, Prop. 308
3. ^ Hobbs, Prop. 310
• Hobbs, C.A. (1921). Solid Geometry. G.H. Kent. pp. 397 ff.