Pt1_Sec 4.5 Circles - Section 4.5 Circles in Euclidean...

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Section 4.5: Circles in Euclidean Geometry (Part 1: Through Corollary 4.5.13)Definition:Given a point P and a real number r > 0 , the circleC( P, r )is the set of points with a common distance r from point P, i.e., the set of points Q in the plane such that PQ = r . Point P is the centerand number r is the radiusof the circle C( P, r ) . Theorem 4.5.1, The "Three Points Circle" Theorem:In the Euclidean Plane, three distinct non-collinear points determine a unique circle. Proof: Let A, B, and C be three non-collinear points. Let l and m be the perpendicular bisectors of ABand BC, resp., intersecting these segments at midpoints M and N, resp. . Suppose lis parallel to m. By Theorem 3.4.8, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other. Therefore, since ABsuuurl, ABsuuurm. Let D be the point where ABintersects m . D is not equal to N or else A, B, and C would be collinear. Thus, BDN has angle sum greater than 180°, which is a impossible in Euclidean Geometry. Therefore, l is not parallel to m. Therefore, land mintersect, say a point P. Since P is on the perpendicular bisectors of segments, PA = PB and PB = PC by Theorem 3.2.8. Let r = PA = PB = PC . Then, all three points, A, B, and C are on the circle C( P, r ). Any other possible circle passing through A, B, and C must have its center on both perpendicular bisectors and so its center must be the point P of intersection of these perpendicular bisectors, so there is no other such circle than C( P, r ) passing through A, B, and C. Q E D Note: This proof provides a method for constructing the center of the circle. lmDMNCBAlmPMNABC
2Facts from Theorems (Proofs left as exercises):1) The "Longest Chord" Theorem: The Diameterof a circle is a chord with length = 2r, which is longer than any non-diameter chord of the circle (Theorem 4.5.3). 2) In a circle C( O, r ) with a diameter ABand a chord PQ, a) The "Chord Bisector" Theorem: ABPQif, and only if, ABbisects PQ(i.e., they intersect at the midpoint of PQ).

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