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Chapter 13. Geometry of the Circle C 13-1 Arcs and Angles (pages 541–542) mArBr 5 60, m/A9O9B9 5 60. Therefore, /AOB > /A9O9B9. Radii of the same circle are congruent, so OA > OB and OrAr > OrBr . By definition, nAOB and nA9O9B9 are isosceles triangles with m/OAB 5 m/OBA 5 60 and m/O9A9B9 5 m/O9B9A9 5 60. Then /OAB > /OBA > /O9A9B9 > /O9B9A9 and nAOB , nA9O9B9 by AA,. Writing About Mathematics 1. Yes. If two lines intersect at the center of a circle, they form four central angles that are also vertical angles. Since vertical angles are congruent, the arcs intercepted by each pair of vertical central angles are also congruent. 2. Yes. Since the degree measure of a circle is 360, the four arcs each measure 90 degrees, and the central angles that intercept the arcs each g 13-2 Arcs and Chords (pages 551–552) g Writing About Mathematics 1. The apothem is perpendicular to the chord, so the triangle formed is a right triangle. Also, the apothem is the distance from the center to the chord, so the length of one leg is 3 inches. The hypotenuse of the triangle is the radius and measures 5 inches. By the Pythagorean Theorem, the length of the other leg is 4 inches, so a 3-4-5 right triangle is formed. However, the apothem intersects the chord at its midpoint, so the length of the chord is 8 inches. 2. No. If the arcs belong to two circles with different radii, then the arcs are not congruent. Developing Skills 3. 3 in. 4. 4.5 cm 5. 1.5 ft 6. 12 mm 7. !6 cm 8. 10 in. 9. 24 ft 10. 14 cm 11. 12.4 mm 12. 2!5 yd 13. radius 5 11, diameter 5 22 14. 5 15. 25 16. 30 17. 12!5 18. 12 19. 5!5 20. OB 5 30, DE 5 60 21. OB 5 30, OC 5 15!3 22. LM Applying Skills measure 90 degrees. Therefore, AC and BD are perpendicular. Developing Skills 3. 35 4. 48 5. 90 6. 140 7. 180 8. 60 9. 75 10. 100 11. 120 12. 170 13. 91 14. 89 15. 91 16. 138 17. 138 18. 133 19. 133 20. 227 21. 180 22. 180 23. 100 24. 110 25. 35 26. 115 27. 115 28. 210 29. 145 30. 145 31. 150 32. 215 Applying Skills 33. In a circle, central angles are congruent if their intercepted arcs are congruent. Since AB > CD , /AOB > /COD. All radii of the same circle are congruent, so OA > OB > OC > OD. Therefore, nAOB > nCOD by SAS. C C 34. /AOC and /BOD are vertical angles, so they are congruent. Radii of the same circle are congruent, so OA > OB > OC > OD. Then nAOC > nBOD by SAS. Since corresponding parts of congruent triangles are congruent, AC > BD. 23. We are given (O > (Or and CD > AB > ArBr. Since congruent circles are circles with congruent radii, OA > OB > OC > OD > OrAr > OrBr, and nCOD > nAOB > nA9O9B9 by SSS. Corresponding parts of congruent triangles are congruent, so /COD > /AOB > /A9O9B9. C 5 90, mADC C 5 270 35. mAC 36. Since AC and BD, the diagonals of ABCD, are perpendicular, ABCD is a rhombus. /AOC and /BOC are right angles since perpendicular lines intersect to form right angles. Radii of the same circle are congruent, so OA > OB > OC. Therefore, nAOB and nBOC are isosceles right triangles and m/ABO 5 m/CBO 5 45. Then m/ABC 5 m/ABO 1 m/CBO 5 90. A rhombus with a right angle is a square, so ABCD is a square. Hands-On Activity In 1–2, results will vary. 3. By the definition of the degree measure of an arc, since mAB 5 60, m/AOB 5 60 and since 24. We are given (O > (Or and CD > AB > ArBr . Since congruent circles are circles with congruent radii, OA > OB > OC > OD > OrAr > OrBr , and nCOD > nAOB > nA9O9B9 by SSS. Corresponding parts of congruent triangles are congruent, so /COD > /AOB > /A9O9B9. In a circle or in congruent circles, if central angles are congruent, then their intercepted arcs are congruent. Therefore, CD > AB > ArBr . C C 333 C C 16. a. 40 17. a. 54 b. 80 b. 48 c. 80 c. 50 d. 40 d. 55 e. 100 e. 50 18. a. 60 19. a. 120 b. 90 b. 120 c. 210 c. 120 d. 45 d. 60 e. 105 e. 60 f. 30 f. 60 Applying Skills 20. a. Since vertical angles are congruent, /LPR > /SPM. If two inscribed angles of a circle intercept the same arc, then they are congruent. Therefore, /RLP > /MSP and nLPR , nSPM by AA,. 25. We are given that diameter AOB intersects chord CD at E and bisects CD at B. Then, CB > BD . In a circle, congruent chords have congruent central angles, so /COE > /DOE. All radii of a circle are congruent, so OC > OD, and OE > OE by the reflexive property of congruence. Therefore, nOEC > nOED by SAS. Corresponding parts of congruent triangles are congruent, so /DEO > /CEO. If two lines intersect to form congruent adjacent angles, then they are perpendicular, so AOB ' CD. A diameter perpendicular to a chord bisects the chord, so E is the midpoint of CD and DE > CE. C C C 26. 5 cm 27. In a circle, if the lengths of two chords are unequal, then the shorter chord is farther from the center. AB is farther from the center than BC, so AB < BC. BC is farther from the center than AC, so BC < AC. By the transitive property of inequality, AB < BC < AC. If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal and the larger angle lies opposite the longer side. Therefore, /B is the largest angle of nABC. b. 8 cm 21. If mAB 5 100 and mBC 5 130, mCA 5 130. Therefore, AB > CA. In a circle, congruent arcs have congruent chords, so AB > CA and nABC is isosceles. C C C C C 22. a. In a parallelogram, opposite angle are congruent. Therefore, /ABC > /ADC or m/ABC 5 m/ADC. The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc, so 13-3 Inscribed Angles and Their Measures (pages 555–558) C C m/ABC 5 12mABC and m/ADC 5 12mADC . Writing About Mathematics C By the substitution postulate, 12mABC 1. Draw diameter AOB. Choose any point C on the circle and draw CA and CB. Inscribed /ACB intercepts a semicircle, so it is a right angle. Triangle ACB is a right triangle with hypotenuse AOB and legs CA and CB. 2. Yes. The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc. Therefore, m/ABC 5 25 and m/PQR 5 25, so /ABC > /PQR. Developing Skills 3. 44 4. 36 5. 85 6. 100 7. 140 8. 24 9. 90 10. 120 11. 190 12. 250 13. a. 160 14. a. 50 15. a. 34 b. 44 b. 100 b. 68 c. 56 c. 50 c. 84 d. 112 d. 80 d. 34 e. 200 e. 160 e. 42 C C C C 5 mADC C 5 180 b. mABC 5 12mADC or mABC 5 mADC . c. By the reasoning in part a, m/BAD 5 m/BCD 5 90 and ABCD is equiangular. An equiangular parallelogram is a rectangle, so ABCD is a rectangle. C C C 23. It is given that mDE 1 mEF 5 mFGD, C C C C C so mDEF 5 mFGD. Since mDEF 1 mFGD 5 360, mDEF 5 180. The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc, so m/DEF 5 90 and nDEF is a right triangle. 24. It is given that AB > CD. Since AOC and BOD are diameters, and /ABC and /BCD are inscribed in semicircles. Therefore, /ABC and /BCD are right angles and congruent. If two inscribed angles of a circle intercept the same arc, then they are congruent, so /BAC > /CDB. Thus, nABC > nDCB by ASA. 334 25. In a circle, congruent chords have congruent arcs. Since AB > CD, AB > CD. Vertical angles are congruent, so /AEB > /DEC. If two inscribed angles of a circle intercept the same arc, then they are congruent, so /BAE > /CDE. Therefore, nABC > nDCB by AAS. 30. In Exercise 26, we proved that a trapezoid inscribed in a circle must be isosceles. Since AB y DC, ABCD is a trapezoid and thus isosceles. It follows that DA > BC. Two chords are congruent if and only if their arcs are congruent. Thus, DA > BC , so CA > CB 1 BA > DA 1 BA > BD . Similarly, CB y ED and DBCE is an isosceles trapezoid. Again, two chords are congruent if and only if their arcs are congruent. Thus, EC > DB > CA . Since EF y DC, DCFE is also an isosceles trapezoid with ED > FC. Thus, DF > DE 1 EF > FC 1 EF > EC , and CA > DB > DF > EC . C C C 26. Let ABCD be inscribed in circle O with AB y CD. Draw AC. If parallel lines are cut by a transversal, then alternate interior angles are congruent, so /ACD > /BAC or m/ACD 5 m/BAC. The measure of an inscribed angle of a circle is equal to one-half the measure of its C C m/BDC 5 12mBC . By the substitution postulate, 1 1 2 mAD 5 2 mBC or mAD 5 mBC . Therefore, C C C C C AD > BC . In a circle, congruent arcs have C C C 13-4 Tangents and Secants (pages 564–567) Writing About Mathematics 1. No. Since l is tangent to circle O at A, l ' OA. Since m is tangent to circle O at B, m ' OB. Both OA and OB are segments of AOB. If two lines are each perpendicular to the same line, then they are parallel. Therefore, l || m and the lines do not intersect. 2. A polygon inscribed in a circle intersects the circle at each of its vertices, whereas a circle inscribed in a polygon intersects each of its sides at exactly one point. Developing Skills C C or m/ACD 5 12mAD and m/BDC 5 12mBC . By C5 C C C C. mAD 5 mBC . Therefore, AD > BC C > BC C , mAD C 5 mBC C . The measure of 28. Since AD 1 2 mBC an inscribed angle of a circle is equal to one-half the measure of its intercepted arc, so m/ACD 5 12mAD and m/BDC 5 12mBC . Halves of equal quantities are equal, so m/ACD 5 m/BDC and /ACD > /BDC. If two lines are cut by a transversal so that the opposite interior angles formed are congruent, then they are parallel. Therefore, AB y CD. C C C 27. We are given that AB y CD. Draw AC. If parallel lines are cut by a transversal, then alternate interior angles are congruent, so /ACD > /BAC or m/ACD 5 m/BAC. The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc, so the substitution C 31. In a circle, two chords are congruent if and only if their arcs are congruent. Since AB is not congruent to CD , AB is not congruent to CD. Since opposite sides of ABCD are not congruent, ABCD is not a parallelogram. congruent chords, so AD > BC and ABCD is an isosceles trapezoid. postulate, 12mAD C C C C C C C C C C C C intercepted arc, so m/ACD 5 12mAD and C C C C C C C 3. AB 5 10, BC 5 15, CA 5 15; since BC 5 CA, BC > CA so nABC is isosceles. 4. AB 5 40, BC 5 50, CA 5 30; since 302 1 402 5 502, nABC is a right triangle. 5. a. 25 6. a. 7 7. a. 24 b. 20 b. 14 b. 16 c. 40 c. 10 c. 36 8. a. 7 9. a. 41 10. a. 12 b. !13 b. 40 b. 12 c. !13 c. 40 c. 5 d. 13 11. a. 3 12. a. 4 d. 8 b. 6 b. 3 e. 11 c. 8 c. 3 f. 12 d. 10 29. In a circle, all radii are congruent, so OA > OB > OC > OD. Vertical angles are congruent, so /AOB > /COD. Then nAOB > nCOD by SAS. Corresponding parts of congruent triangles are congruent, so /BAC > /DCA. If two lines are cut by a transversal so that the opposite interior angles formed are congruent, then they are parallel. Therefore, AB y CD. 335 13. a. !3 cm b. 1.73 cm 14. a. 80 b. 70 c. 50 e. 80 f. 10 g. 20 i. 110 j. 130 k. 360 Applying Skills g perpendicular to a radius at a point of tangency, so AC ' OA and m/OAC 5 90. The sum of the measures of the angles in a triangle is 180, so m/AOC 5 45. Since /OAC > /AOC, AC > AO. d. 160 h. 100 g 15. PQ and PR are tangent to circle O at Q and R, respectively. Tangent segments drawn to a circle from an external point are congruent, so PQ > PR. PO > PO by the reflexive property of congruence. Radii of a circle are congruent, so OQ > OR. Therefore, nOPQ > nOPR by SSS. Corresponding parts of congruent triangles are congruent, so /OPQ > /OPR. By definition, b. Since nAOC is a 45-45-degree right triangle, OC 5 "2OA. c. From part a, m/ACB 5 90 and AC > AO. Radii of a circle are congruent, so OA > OB. By the same reasoning in part a, BC > BO. By the transitive property of congruence, AO > OB > BC > AC. Since AOBC is equilateral and has a right angle, it is a square. h PO bisects /RPQ. g g 19. Isosceles nABC with vertex A is circumscribed about circle O, and D, E, and F are the points of tangency of AB, BC, and AC, respectively. By the isosceles triangle theorem, /DBE > /FCE. If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents. Therefore, m/OBE 5 12m/DBE and m/OCE 5 12m/FCE, so /OBE > /OCE. Since BC is tangent to circle O at E, BC ' OE, forming congruent right angles /BEO and /CEO. OE > OE by the reflexive property of congruence, so nBEO > nCEO by AAS. Corresponding parts of congruent triangles are congruent, so BE > CE and E is the midpoint of BC. 16. PQ and PR are tangent to circle O at Q and R, respectively. Tangent segments drawn to a circle from an external point are congruent, so PQ > PR. PO > PO by the reflexive property of congruence. Radii of a circle are congruent, so OQ > OR. Therefore, nOPQ > nOPR by SSS. Corresponding parts of congruent triangles are congruent, so /QOP > /ROP. By definition, h OP bisects /QOR. 17. a. Tangent segments drawn to a circle from an external point are congruent, so PQ > PR. By the isosceles triangle theorem, /PQR > /PRQ. b. From part a, PQ > PR and /PQR > /PRQ. SP > SP by the reflexive property of congruence, so nQPS > nRPS by SAS. Corresponding parts of congruent triangles are congruent, so QS > RS and QS 5 RS. Also, /QSP > /RSP. If two lines intersect to form congruent adjacent angles, then they are perpendicular. Therefore, OP ' QR. 20. a. If a line is tangent to a circle, then it is perpendicular to a radius at a point on the g g circle. Therefore, AB ' OA and AB ' OrB. If two lines are each perpendicular to the same line, then they are parallel, so OA y OrB. We are given that OA < O9B. Since one pair of opposite sides of OABO9 is not both congruent and parallel, OABO9 is not a c. The length of the altitude to the hypotenuse of a right triangle is the mean proportional between the lengths of the segments of the hypotenuse. Thus, if x 5 OS, then x4 5 10 42 x or 2x2110x 5 16. Solving for x gives x 5 2 or 8. Since OS < SP, OS 5 2 and SP 5 8. 18. a. We are given perpendicular tangents AC and BC. Since AC ' BC, m/ACB 5 90. If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents. Therefore, m/ACO 5 45. A tangent line is g parallelogram. Therefore, OOr cannot be g parallel to AB. g g b. From part a, AB ' OA and AB ' OrB. Therefore, /OAC and /O9BC are right angles and congruent. By the reflexive property of congruence, /C > /C. Therefore, nOAC , nO9BC by AA,. c. AC 5 8, AB 5 4, OC 5 2!17, O9C 5 3!17, OO9 5 "17 336 2. Yes. Vertical angles are congruent, and each central angle is equal to the measure of its intercepted arc. One-half of the sum of two equal intercepted arcs is the intercepted arc. Developing Skills 3. 35 4. 30 5. 30 6. 166 7. 170 8. 60 9. 50 10. 45 11. 50 12. 130 13. 130 14. 180 15. 20 16. 100 17. 80 18. 30 19. 60 21. a. We are given that OA 5 O9B, so OA > OrB. If a line is tangent to a circle, then it is perpendicular to a radius at that point on the g g circle. Therefore, AB ' OA and AB ' OrB, forming congruent right angles /OAC and /O9BC. Vertical angles are congruent, so /OCA > /O9CB. nOCA > nO9CB by AAS. Corresponding parts of congruent triangles are congruent, so OC > OrC and OC 5 O9C. b. From part a, nOCA > nO9CB. Corresponding parts of congruent triangles are congruent, so AC > BC and AC 5 BC. Hands-On Activity a. Results will vary. b. (1) By definition, the center, P, is the intersection of the bisectors of the angles of a regular polygon. If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle. Thus, the center P is equidistant from the sides of the regular polygon. The distance from a point to a line is equal to the length of the perpendicular from the foot. Thus, the apothems are all congruent. C C 28. a. 150 d. 30 Applying Skills b. 150 e. 75 20. 21. 24. 27. mRQ 5 135, mRSQ 5 225 55 22. 140 23. 35 140 25. 70 26. 220 a. 125 b. 55 c. 55 d. 90 e. 90 f. 125 g. m/AED 5 12 (55 1 125) 5 90. Therefore, BD ' AC. A line through the center of a circle that is perpendicular to a chord bisects the chord, so BD bisects AC. c. 30 f. 105 g 29. a. We are given that AB y CP. If two parallel lines are cut by a transversal, then the alternate interior angles formed are congruent. Therefore, /OCP > /OEA and /OPC > /OAE, and nOPC , nOAE by AA,. (2) Each circle constructed in part a used one of the apothems as the radius. Since the apothems are all congruent, every apothem is also a radius of the circle, so the foot of each apothem is on the circle. (3) By construction, each side of the regular polygon is perpendicular to one of the apothems at the foot of the apothem. If a line is perpendicular to a radius at a point on a circle, then it is tangent to the circle. Since the foot of each apothem is on the circle and each apothem is a radius of the circle, the sides of the regular polygon are tangent to the circle. C C C C C b. mAD 5 75, mCF 5 45, mFB 5 60, mBD 5 75, mAC 5 105, m/P 5 30 30. Vertical central angles are congruent and intercept congruent arcs. Therefore, /FOE > C C /GOD and mEF 5 mDG . We are given that C C C 5 mFB C and mDGB C C 2 mEF C 2 mDG mEFB C . Thus, by substitution, mFB C 5 mEFB C2 5 mGB C C C C mEF 5 mDGB 2 mDG 5 mDG . The measure mEFB 5 mDGB . By the partition postulate, 2 c. r2 5 a2 1 A 2s B 13-5 Angles Formed By Tangents, Chords, and Secants (pages 572–574) of an angle formed by a tangent and a secant is Writing About Mathematics cepted arcs, so m/A 5 21 (mDGB 2 mFB ) and m/C 5 12 (mEFB 2 mGB ) . By substitu- equal to one-half the difference of the inter- C g 1. Yes. Let AB be tangent to the circle O at A. Then g g AB ' OA and m/AOB 5 90. Let AC be a secant that intersects circle O at A and C. Then m/OAB 5 m/OAC 1 m/CAB. Therefore, m/OAC Þ 90 and the radius OA is not tion, m/A 5 C C C C) 5 C 2 mFB 1 2 (mDGB C ) 5 m/C, so /A > /C. C 2 mGB 1 2 (mEFB Therefore, OA > OC and nAOC is isosceles. g perpendicular to the secant AC. 337 C 12. 15. 18. 19. 20. 21. 22. 23. 31. Let D be a point on major ADB . Since C 5 120 and mADB C 5 240. m/AOB 5 120, mAB The measure of an angle formed by two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs, so m/P 5 12 (240 2 120) 5 60. Tangent segments drawn to a circle from an external point are congruent, so /PAB > /PBA. Since the sum of the measures of the angles of a triangle is 180, m/PAB 5 m/PBA 5 60. Therefore, nABP is equiangular. If a triangle is equiangular, then it is equilateral, so nABP is equilateral. 27 13. 20 and 6 16 16. 6 8 AC 5 24, AB 5 6, BC 5 18 AC 5 20, AB 5 5, BC 5 15 AC 5 16, AB 5 4, BC 5 12 AC 5 25, AB 5 9, BC 5 16 20 24. 12 26. AB 5 27. 29. 31. , 3 $21 5 AC 5 8 28. 8!2 AD 5 3, AE 5 6 14. 14 and 2 17. 4 25. 30 15 $21 5 30. AD 5 7, AE 5 12 15 2 13-7 Circles in the Coordinate Plane (pages 584–587) g 32. We are given secant ABC intersecting circle O at A and B and chord BD. Assume that m/CBD 5 Writing About Mathematics 1. Yes. Consider a circle with center (h, k) and radius r. Two points on the circle are (h 2 r, k) and (h 1 r, k). The chord connecting these points goes through the center, so it is a diameter. These points have the same y-coordinate, so this diameter is horizontal. Two other points on the circle are (h, k 1 r) and (h, k 2 r). The chord connecting these points goes through the center, so it is a diameter. These points have the same y-coordinate, so this diameter is vertical. 2. Yes. Dividing each side by 3 gives x2 1 y2 5 4. This is the equation of a circle centered at the origin with a radius of 2. Developing Skills 3. x2 1 y2 5 9 4. (x 2 1) 2 1 (y 2 3) 2 5 9 5. (x 1 2) 2 1 y2 5 36 6. (x 2 4) 2 1 (y 1 2) 2 5 100 7. (x 2 6) 2 1 y2 5 81 8. (x 1 3) 2 1 (y 1 3) 2 5 4 9. x2 1 y2 5 4 10. x2 1 y2 5 16 11. (x 2 2) 2 1 (y 2 9) 2 5 16 12. (x 1 1) 2 1 (y 2 3) 2 5 16 13. x2 1 (y 2 12) 2 5 25 14. (x 1 6) 2 1 (y 2 1) 2 5 9 15. (x 2 1) 2 1 (y 2 6.5) 2 5 76.25 16. (x 2 10) 2 1 (y 1 1) 2 5 73 17. (x 1 3) 2 1 (y 2 2) 2 5 4 18. (x 2 4) 2 1 (y 2 3) 2 5 16 19. (x 2 5) 2 1 (y 1 5) 2 5 9 20. (x 1 4) 2 1 (y 1 4) 2 5 25 21. (x 1 1) 2 1 y2 5 1 22. x2 1 (y 1 2) 2 5 36 C C 1 2 mBD . /ABD is an inscribed angle, so m/ABD 5 12mAD . Since /ABD and /CBD form a linear pair, m/CBD 1 m/ABD 5 180. By the C C C 1 mAD C 5 360. However, mBD C1 or mBD C C C C 5 360. mAD 2 360 since mBD 1 mAD 1 mAB C. Our assumption is false and m/CBD Þ 12mBD substitution postulate, 12mBD 1 12mAD 5 180 13-6 Measures of Tangent Segments, Chords, and Secant Segments (pages 579–581) Writing About Mathematics 1. Yes. If two chords intersect, the product of the measures of the segments of one chord is equal to the product of the segments of the other. Since AB 5 24, and M is the midpoint of AB, AM 5 BM 5 24. Because 12 3 12 5 144, any chord of circle O that intersects AB at M is separated by M into two segments such that the product of the lengths is also 144. 2. No. If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. Since (AP)(BP) 5 (CP)(DP) and AP > CP, BP must be less than CP. Developing Skills 3. 8 4. 6 5. 10 6. 7 7. 8 8. 28 9. 4 10. 8 11. 4 338 23. 28. y 1 O 1 y x (2, 25) (1, 1) 1 x O 1 24. y (24, 4) O 25. 1 29. (x 2 2) 2 1 (y 2 3) 2 5 169 30. Yes. The equation of the circle is x2 1 y2 5 32. Substituting (4, 4) into the equation gives 42 1 42 5 32. Therefore, (4, 4) is a point on the circle. 31. Yes. It is equivalent to (x 1 2) 2 1 (y 2 1) 2 5 25, which is the equation for a circle centered at (22, 1) with a radius of 5. Applying Skills 32. a. y 5 2x 1 2 b. x 5 0 1 c. y 5 3x 1 2 d. (0, 2) x 1 y (232 , 1) x O 21 1 26. y O1 21 27. e. The perpendicular bisector of a side is equidistant from the endpoints of each side, so the circumcenter, which is the intersection of the three perpendicular bisectors, is equidistant from the vertices of nABC. f. Yes. Let (0, 2) be the center of the circle and A(2, 6) be a point on the circle. Then the radius is !20 and the equation of the circle is x2 1 (y 2 2)2 5 20. Substituting B(24, 0) and into the equation gives 242 1 (0 2 2) 2 5 16 1 4 5 20, so B is on the circle. Substituting C(4, 0) into the equation gives 42 1 (0 2 2)2 5 16 1 4 5 20, so C is on the circle. 33. a. Yes. The x-coordinate of the incenter is 3. 6(3) 1 8y 5 10 x ( 5 , 23 2 4 ) y 18 1 8y 5 10 1 O (0, 0) 1 y 5 21 x The incenter is at (3, 21). b. The incenter is on each of the angle bisectors, and a point on the bisector of an angle is equidistant from the sides of the angle. c. Yes. The equation of PR is y 5 2, so (3, 2) is on PR. Since the incenter of the triangle is 339 12. a. (5, 5) b. Tangent 13. a. (2, 2) b. Tangent 14. a. (0, 24) and (2, 22) b. Secant 15. x 5 3 16. y 5 24 17. y 5 x 2 4 18. y 5 2x 2 10 Applying Skills 19. a. y 5 3x 2 5 b. x 5 0 c. P(0, 25) d. PB 5 0 and PE 5 0, so (PA)(PB) 5 (PD)(PE) 5 0. 20. a. y 5 x 1 2 b. y 5 10 c. P(8, 10) d. PA 5 2!2, PB 5 16 !2, PD 5 8 ? (PA)(PB) 5 (PD) 2 (2 !2)(16 !2) 5 82 64 5 64 ✔ 21. a. y 5 2x 1 6 b. y 5 x 2 6 c. (6, 0) d. PA 5 PB 5 18 22. The line and the circle intersect at a single point: (18, 24). 23. a. Both points are solutions to the equation (x 2 2) 2 1 (y 2 3) 2 5 25. b. 4 24. a. The equation of AB is y 5 234x 1 23 . The equation of BC is x 5 2. The equation of AC is equidistant from sides of the circle, PQ and QR also intersect the circle at one point. d. (x 2 3) 2 1 (y 1 1) 2 5 9 34. The equation of the perpendicular bisector of AB is y 5 22x 2 4 and the equation of the perpendicular bisector of BC is y 5 21. 21 5 22x 2 4 2x 5 23 x 5 232 The circumcenter of nABC is (21.5, 21), which is also the center of the circle. The radius equals the distance between (21.5, 21) and A(21, 3) 5 "(21 1 1.5) 2 1 (3 1 1) 2 5 !16.25. The equation of the circle is (x 1 1.5) 2 1 (y 1 1) 2 5 16.25. 35. a. d. 36. a. b. c. d. 12 !3 ft b. (s, s !3) c. A s, s !3 3 B 6 ft e. 12 ft (x 2 13)2 1 (y 2 13)2 5 169 (8, 1) and (8, 25); (18, 1) and (18, 25) (8, 13) or (18, 13) 1,800 ft and 800 ft y 5 10. AB intersects the circle at (2325, 515) . 13-8 Tangents and Secants in the Coordinate Plane (pages 592–593) BC intersects the circle at (2, 7). AC intersects the circle at (21, 10). b. The sides of the triangle each intersect the circle at exactly one point. c. Yes. Each side of the triangle is tangent to the circle. Writing About Mathematics 1. Yes. The equation of the circle is (x 2 r)2 1 (y 2 k)2 5 r2. The equation of the y-axis is x 5 0. Solving for y gives: (0 2 r) 2 1 (y 2 k) 2 5 r2 r2 1 (y 2 k) 2 5 r2 Review Exercises (pages 598–599) (y 2 k) 2 5 0 1. a. 100 2. a. 100 3. a. 40 b. 50 b. 110 b. 130 c. 80 c. 75 c. 65 d. 90 d. 30 d. 50 e. 90 e. 100 e. 95 4. PB 5 6, BC 5 18 5. 10 6. AE 5 3, EC 5 8 7. (1) 8. (3) 9. 60 10. 15 cm 11. AB 5 4, AC 5 12 12. mAB 5 mAC 5 100, mBC 5 160 13. Let ABCD be inscribed in circle O with AB y CD. Draw AC. If parallel lines are cut by a transversal, then alternate interior angles are congruent, so /ACD > /BAC or m/ACD 5 m/BAC. The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc, so m/ACD 5 12mAD and m/BDC 5 12mBC . By y5k Thus, the y-axis intersects the circle in the point (0, k), so it is tangent to the circle. g 2. No. Since the slope of AB is not the negative reciprocal of the slope of CA, it is not perpendicular to CA at the point of intersection. g Therefore, AB is not tangent at A. Developing Skills 3. a. (0, 6) b. Tangent 4. a. (6, 8) and (8, 6) b. Secant 5. a. (3, 4) and (4, 3) b. Secant 6. a. (1, 3) and (21, 23) b. Secant 7. a. (0, 23) and (3, 0) b. Secant 8. a. (2, 2) and (22, 22) b. Secant 9. a. (4, 3) and (23, 24) b. Secant 10. a. (2, 4) and (4, 2) b. Secant 11. a. (23, 3) b. Tangent C C C C 340 C C C C C the substitution postulate, 12mAD 5 12mBC or mAD 5 mBC . Therefore, AD > BC . In a circle, congruent arcs have congruent chords, so AD > BC and ABCD is an isosceles trapezoid. C C (PAB)(PA) 5 (PCD)(PD). By the division postulate, PA 5 PD. Then, by the subtraction postulate, PAB 2 PA 5 PCD 2 PD, so AB 5 CD and AB > CD. Two chords are equidistant from the center of a circle if and only if the chords are congruent. Therefore, AB and CD are equidistant from the center of the circle. 14. a. Since AB y CD, alternate interior angles /B and /C are congruent so m/B 5 m/C. m/B 5 12mAC and m/C 5 12mBD . By the C C C 5 mBD C . If two substitution postulate, mAC 18. 12 cm inscribed angles of a circle intercept the same arc, then they are congruent. Therefore, /A > /C and /D > /B. By the transitive property of congruence, /A > /B and /C > /D. If two angles of a triangle are congruent, then the sides opposite these angles are congruent, so AE > BE and CE > DE. Therefore, nABE and nCDE are isosceles. C Exploration (pages 599–600) a. Place the point of your compass on a point P on a g line AB. With any convenient radius, draw arcs h C b. From part a, mAC 5 mBD . Therefore, AC > BD . C C b. The interior angles of a regular hexagon each measure 120°. The angle bisectors of the angles of the polygon intersect in the center of the polygon and form congruent equilateral triangles. Use the construction from part a to construct equilateral nABP. Then, construct equilateral triangle nBCP. Next, construct equilateral triangle nCDP. Continue this process until you have constructed equilateral nFAP. Since the equilateral triangles all share two sides, they are all congruent. Thus, the sides of the resulting hexagon, ABCDEF, are all congruent. Each interior angle of the hexagon is the union of two angles of two adjacent equilateral triangles. Since the angles of an equilateral all measure 60°, each interior angle of the hexagon measures 60 1 60 5 120°. Therefore, the hexagon formed is a regular hexagon. c. From part a, /A > /C and /D > /B. Therefore, nABE , nCDE by AA,. 15. We are given that AB > BC > CD > DA. Therefore, mAB 5 mBC 5 mCD 5 mDA 5 90. The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc, so m/A 5 12 (mBC 1 mCD ) 5 12 (180) 5 90. Therefore, m/A is a right angle. In a circle, if two arcs are congruent, then their chords are congruent. Therefore, AB > BC > CD > DA and ABCD is a rhombus. A rhombus with a right angle is a square, so ABCD is a square. C C C C C C C h that intersect PA at C and PB at D. With C and D as centers and the compass open to a radius equal to CD, draw arcs that intersect at a point E. Connect C, D, and E to form nCDE. Congruent radii were used to draw CD, DE, and CE, so CD > DE > CE and nCDE is equilateral. C C C 16. Let ABC be a triangle with right angle /B and M the midpoint of AC. Then, MA > MC. Let nABC be inscribed in a circle. Since a right angle C is inscribed in a semicircle, ABC is a semicircle and AC is a diameter. Since MA > MC, M is the center of the circle. All radii of the same circle are congruent, so MA > MC > MB and MA 5 MC 5 MB, and the midpoint of the hypotenuse is equidistant from the vertices of the triangle. c. The diagonals of a square bisect each other, are perpendicular, and are congruent. Let O be the center of the circle. Draw a diameter AOB of the circle. Construct the perpendicular bisector of the diameter. Let C and D be the points where the perpendicular bisector intersects the circle. Since AO and OB are radii, AO > OB, and the perpendicular bisector DC passes through O. But OD and OC are also radii. Therefore, AB and CD bisect each other, are congruent, and are perpendicular, so ACBD is a square. 17. Since PAB > PCD, PAB 5 PCD. If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. Therefore, 341 d. From the construction in part c, AC , CB , BD , and DA are all congruent since their central angles are all right angles. Thus, the arcs formed by bisecting these arcs are all congruent (since halves of congruent arcs are congruent). Two chords are congruent if and only if their arcs are congruent. Therefore, the chords formed by the arcs are all congruent, so the sides of the polygon formed are all congruent. Draw segments from the vertices of the polygon to the center of the circle. These segments are all congruent because they are radii of the same circle. Thus, congruent isosceles triangles are formed by SSS. The base angles of these triangles are congruent, so it follows that the interior angles of the octagon are all congruent. Therefore, the octagon formed is a regular octagon. Part III 13. Statements f. Construct six congruent and adjacent equilateral triangles with the center of the circle as a vertex and using two radii as sides. 14. CC C C Reasons g g g g 1. EF ' AB and 1. Given. EF ' CD 2. /AEF and /CEF are right angles. 3. /AEF > /CEF 4. EF > EF 5. EA > EC 6. nAEF > nCEF 7. FA > FC 2. Definition of perpendicular lines. 3. Right angles are congruent. 4. Reflexive property of congruence. 5. Given. 6. SAS. 7. Corresponding parts of congruent triangles are congruent. x2 1 (x 2 7) 2 5 (x 1 2) 2 x2 1 x2 2 14x 1 49 5 x2 1 4x 1 4 x2 2 18x 1 45 5 0 Cumulative Review (pages 600–603) (x 2 15)(x 2 3) 5 0 Part I 1. 3 2. 1 3. 4 4. 3 5. 1 6. 3 7. 3 8. 2 9. 1 10. 3 Part II 11. a. (2, 2) b. (22, 0) c. Yes. The slope of DE is 12 and the slope of AC x 5 15 u x 5 3 x cannot be 3 since 3 2 7 5 24. The lengths of the sides of the triangle are 15, 8, and 17. Part IV 15. a. (25, 2); ry=x (5, 2) 5 (2, 5), R90 (2, 5) 5 (25, 2) b. ry-axis c. No. Rotation is a direct isometry and line reflection is an opposite isometry. is 12 . d. Yes. By the reflexive property of congruence, /B > /B. If two parallel lines are cut by a transversal, then the corresponding angles are congruent. Since DE y AC, /CAD > /EDB and nABC , nDBE by AA,. 12. Statements 1. ABCD with AC > BD, and AC and BD bisect each other. 2. ABCD is a parallelogram. 3. ABCD is a rectangle. 16. a. We are given that AD : DC 5 1 : 3 and BE : EC 5 1 : 3. Let AD 5 x, so DC 5 3x. Let BE 5 y, so EC 5 3y. AC 5 AD 1 DC 5 4x and AC 4 5 4x BC 5 BE 1 EC 5 4y. DC 3x 5 3 and BC EC 4y 5 3y 5 43 . Therefore, AC : DC 5 BC : EC. b. From part a, AC : DC 5 BC : EC. By the reflexive property of congruence, /C > /C. Therefore, nABC , nDEC by SAS,. Reasons 1. Given. 2. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 3. If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. 342 Chapter 14. Locus and Construction h 14-1 Constructing Parallel Lines (pages 607–609) radius, mark off point Y on BD. Draw XY. ABYX is a square. Writing About Mathematics 1. The distance between two parallel lines is defined as the length of the perpendicular from any point on one line to the other line. Therefore, since CE is the perpendicular with length 2PQ, g c. Use Construction 1 to construct a segment AB congruent to the segment with length a. Use Construction 2 to construct an angle congruent h to /A on AB . On the side of this angle not containing point B, use Construction 1 to construct a segment AC congruent to the segment with length b. Use Construction 7 to construct a line parallel to AB through point C. On this line, use Construction 1 to construct CD congruent to the segment with length a such that CD is on the same side of AC. Draw BD. ABDC is a parallelogram. g every point on CD is at a distance 2PQ from AB. 2. If two coplanar lines are each perpendicular to the same line, then they are parallel. Thus, g g g g GH y AB and GH y CD. The distance between two parallel lines is defined as the length of the perpendicular from any point on one line to the other line. Thus, the distance from any point on g g g GH to AB and CD is 12GH. Developing Skills 3. a. Use Construction 7. b. Mark any point Q on l. Use Construction 6 to construct the perpendicular to m through Q. Let R be the intersection of this perpendicular and line m. Then use Construction 3 to construct the perpendicular bisector of QR. c. Construct a line parallel to l (and above l) that 5. h is the same distance away as m. Extend QR to a point S such that QR 5 RS. Then use Construction 7 to construct a line parallel to l through S. d. Yes. If two of three lines in the same plane are each parallel to the third line, then they are parallel to each other. 4. a. Use Construction 1 to construct a segment AB congruent to the segment with length b. Use g 6. g Construction 5 to construct AC and BD perpendicular to AB with C and D on the same side of AB. Set the compass radius to a. 7. h With A as the center, mark off point X on AC . With B as the center and using the same h radius, mark off point Y on BD. Draw XY. ABYX is a rectangle. b. Use Construction 1 to construct a segment AB congruent to the segment with length a. g 8. g Use Construction 5 to construct AC and BD perpendicular to AB with C and D on the same side of AB. Set the compass radius to a. h With A as the center, mark off point X on AC . With B as the center and using the same 343 d. Repeat part c, but construct AC to be congruent to the segment with length a. ABDC is a rhombus. a. Construct the perpendicular bisector of AB at C. Construct the perpendicular bisector of AC at D and of CB at E. Then D, C, and E divide AB into four congruent parts. b. Set the compass radius to AB. Draw a circle. c. Set the compass radius to AC. Draw a circle. a. Use Construction 7. b. Use Construction 3 to find the midpoint M of AC. Draw BM. Then BM is the median to AC. c. Use Construction 3 to find the midpoint N of BC. Draw AN. Then AN is the median to BC. d. Yes. Any two medians of a triangle intersect in the same point, and two points determine a line. Thus, the line drawn from C to the median of AB is the same as that drawn from C to P. a. Use Construction 3. b. Use Construction 3. c. Use Construction 3 to find the midpoint M of AB. d. P and M. Any two perpendicular bisectors of the sides of a triangle intersect in the same point. Thus, the line containing P and M is the perpendicular bisector of AB. a. Use Construction 6 to construct a line perpendicular to AC through B intersecting AC at D. BD is the altitude to AC. b. Use Construction 6 to construct a line perpendicular to BC through A intersecting BC at E. AE is the altitude to BC.