Dividing a circle into areas "F" can also be derived, which will solve the problem. Its vertices include the "n" points on the circle, referred to as the exterior vertices, as well as the interior vertices, the intersections of distinct chords in the interior of the circle. The "generic intersection" assumption made above guarantees that each interior vertex is the intersection of no more than two chords. Thus the main task in determining "V" is finding the number of interior vertices. As a consequence of the lemma, any two intersecting chords will uniquely determine an interior vertex. These chords are in turn uniquely determined by
Circle of a sphere there is exactly one circle that can be drawn through three given points. The proof can be extended to show that the points on a circle are all a common angular distance from one of its poles. To show that a non-trivial intersection of two spheres is a circle, assume (without loss of generality) that one sphere (with radius formula_1) is centered at the origin. Points on this sphere satisfy Also without loss of generality, assume that the second sphere, with radius formula_3, is centered at a point on the positive x-axis, at distance formula_4 from the origin. Its points