AMSCO'S Geometry. New York - Rye High School

Spheres 461Proof Statements Reasons1. Draw a line through O, 1. Through a given point there is oneperpendicular to plane p at C. line perpendicular to a given plane.2. OCA and OCB are 2. A line perpendicular to a plane isright angles.perpendicular to every line in theplane through the intersection ofthe line and the plane.3. OA > OB3. A sphere is the set of points inspace equidistant from a fixedpoint.4. OC > OC4. Reflexive property.5. OAC OBC 5. HL.6. CA > CB6. Corresponding sides of congruenttriangles are congruent.7. The intersection is a circle. 7. A circle is the set of all points in aplane equidistant from a fixedpoint.We can write Theorems 11.14a and 11.14b as a single theorem.Theorem 11.14The intersection of a plane and a sphere is a circle.In the proof of Theorem 11.14b, we drew right triangle OAC with OA theradius of the sphere and AC the radius of the circle at which the plane and thesphere intersect. Since OCA is the right angle, it is the largest angle of OACand OA AC. Therefore, a great circle, whose radius is equal to the radius ofthe sphere, is larger than any other circle that can be drawn on the sphere. Wehave just proved the following corollary:Corollary 11.14aA great circle is the largest circle that can be drawn on a sphere.Let p and q be any two planes that intersect the sphere with center at O.Inthe proof of Theorem 11.14b, the radius of the circle is the length of a leg of aright triangle whose hypotenuse is the radius of the sphere and whose other legis the distance from the center of the circle to the plane.This suggests that if twoplanes are equidistant from the center of a sphere, they intersect the sphere incongruent circles.