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678. DEF. A small circle of a sphere is a section made by a plane not passing through the center.
679. DEF. The axis of a circle of a sphere is the diameter perpendicular to the plane of the circle; its ends are the poles of the circle.
680. Cor. 2. The axis of a circle passes through the center of the circle, and conversely.
681. Cor. 3. All great circles of a sphere are equal.
682. COR. 4. Any two great circles of a sphere bisect each other.
For since the plane of each contains the center of the sphere, their intersection is a diameter and bisects both circles.
683. Cor. 5. Every great circle bisects the sphere.
684. Cor. 6. One and only one circle may be drawn through any three points in the surface of a sphere. (A plane is determined by three points.)
685. Cor. 7. A great circle may be drawn through two points B and C in the surface of a sphere. (Three points, B, C, and the center 0, determine a plane.)
Generally there is only one great circle which passes through two given points, but if the given points are ends of a diameter, any number of great circles can be passed through these points.
686. DEF. The distance between two points on the surface of a sphere is the length of the minor arc of a great circle between them.
Ex. 1169. What is the radius of a small circle, if the distance of its plane from the center of the sphere is 9 in., and the radius of the sphere is 15 in.
PROPOSITION II. THEOREM
687. All points in the circumference of a circle of a sphere are equidistant from a pole of the circle.
Hyp. P and P are the poles of circle ABC of a sphere.
= arc PB= arc PC, arc P'A=
= arc P'B= arc P'C.
Hint. — Prove by means of (473) the equality of straight lines PA, PB, and PC.
688. DEF. The polar distance of a circle of a sphere is the distance of a point in the circumference from the nearer pole.
689. SCHOLIUM. A quadrant in Spherical Geometry is the fourth part of the circumference of a great circle.
690. Cor. The polar distance of a great circle is a quadrant.
Ex. 1170. The polar distance of a circle of a sphere is 60°, and the radius of the sphere is 13 in. Find
(a) the distance of its plane from the center.
PROPOSITION III. THEOREM
691. On the surface of a sphere, a point at a quadrant's distance from two other points, not the extremities of a diameter, is the pole of a great circle passing through these two points.
Hyp. P, A, and B are three points on the surface of a sphere, and PA and PB are quadrants.
To prove P is the pole of a great circle AB.
Hint. — Prove that a line OP is perpendicular to plane ABO by means of (461)
692. SCHOLIUM. Theorem III enables us to construct a great circle through two points on the surface of a material sphere by means of the compasses.
From the given points A and B as centers draw arcs with a radius equal to the chord of a quadrant, intersecting in P. From P, with the same radius, draw a circle which is the required one.
693. DEF. A plane is tangent to a sphere when it has one and only one point common with the surface of a sphere.
694. DEF. A line is tangent to a sphere when it has one and only one point common with the surface of a sphere.
695. DEF. Two spheres are tangent if their surfaces have one and only one point common.
696. DEF. A sphere is inscribed in a polyedron if all the faces of the polyedron are tangent to the sphere.
697. DEF. A sphere is circumscribed about a polyedron if all the vertices of the polyedron lie in the surface of the sphere.
PROPOSITION IV. THEOREM
698. A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere.
Нур. . Plane MN is I to radius OA of sphere o at its extremity A.
To prove MN is tangent to the sphere.
(476) Hence B lies without the sphere.
Therefore, every point of MN, except A, lies without the sphere, and MN is tangent to the sphere.
699. Cor. 1. Any straight line perpendicular to the radius of a sphere at its extremity is a tangent to the sphere.
700. CoR. 2. Any line or plane tangent to a sphere is perpendicular to the radius drawn to the point of contact.
PROPOSITION V. THEOREM
701. A sphere may be circumscribed about any tetraedron.
Hyp. ABCD is a tetraedron.
Proof. Let E and F be respectively the centers of the circumscribed circles of A ADC and BCD.
Draw EHI to plane ABC, and FI I to plane BCD. Join K, the midpoint of DC, to F and E.
and in like manner IF lies in this plane.