4. 2 common tangents 5. 1 common tangent common tangents 7. CE 2 0 CD 2 1 DE 2

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1 hapter 0 opright b Mcougal Littell, a division of Houghton Mifflin ompan. Prerequisite Skills (p. 648). Two similar triangles have congruent corresponding angles and proportional corresponding sides.. Two angles whose sides form two pairs of opposite ras are called vertical angles. 3. The interior of an angle is all of the points between the sides of the angle. 4. c? a b 5. c? a b 0.9? ? 0.8? ? < 89 > 65 The triangle is acute. The triangle is obtuse. 6. c? a b ? ? The triangle is right. 8. (8 ) ( ) Lesson Investigating ctivit 0. (p. 650). nswers will var.. Tangent segments from a common eternal point are congruent. 3. MQ 5 MP MN 5 LM 5 7 LQ 5 LM MQ PN 5 PM MN ecause 5 E and 5, it follows that 5 E which makes > E. 0. Guided Practice (pp ). G is a chord because its endpoints are on the circle. is a radius because is the center and is a point on the circle.. tangent E and a tangent segment is. 3. The radius of ( is 3 units. The diameter of ( is 6 units. The radius of ( is units. The diameter of ( is 4 units. 4. common tangents 5. common tangent 6. 0 common tangents 7. E 0 E (3 ) Yes, E is tangent to (. 8. QT 5 QS ST (r 8) 5 r 4 r 36r 34 5 r r 5 5 r Eercises (pp ) Skill Practice. The points and are on (. If is a point on, then is a diameter.. When referring to a segment, a radius and a diameter is used. When referring to a length, the radius and the diameter is used. 3. G; is a point of tangenc. 4. H; is a common tangent. 5. ; is a chord. 6. is a secant. 7. E is a tangent. 8. ; G is the center. 9. ; is a radius. 0. ; is a diameter.. The error is that is not a secant, but rather a chord. The length of chord is 6 units.. The radius of ( is 9 units. The diameter of ( is 8 units. 3. The radius of ( is 6 units. 4. The diameter of ( is units

2 hapter 0, continued 5. 4 common tangents 8. The common tangents are internal because the intersect the segment that joins the centers of the circles. 9. ; Using Theorem 0., RS QR. raw a congruent, parallel segment XP as shown. 5 Q R S 3 P 6. 0 common tangents 7. common tangent 8. Use the onverse of the Pthagorean Theorem. ecause , n is a right triangle and. Theorem 0., is tangent to (. 9. Use the onverse of the Pthagorean Theorem. ecause 9 5 Þ 8, is not perpendicular to and is not tangent to (. 0. The diameter of ( is 0. Using the onverse of the Pthagorean Theorem, and the triangle is a right triangle. This implies that. Using Theorem 0., is tangent to (.. (r 6) 5 r 4 r 3r 56 5 r 576 3r 5 30 r 5 0. (r 6) 5 r 9 r r 36 5 r 8 r 5 45 r (r 7) 5 r 4 r 4r 49 5 r 96 4r 5 47 r The common tangents are eternal because the do not intersect the segment that joins the centers of the two circles. Using Pthagorean Theorem, (XQ) (XP) 5 (QP), (XP) 5 8. XP 5 Ï 5, so RS 5 Ï Using Theorem 0., P > P and P > P. Using the Transitive Propert of Segment ongruence, P > P > P. 3. Sample nswer: Two lines tangent to the same circle will not intersect when the lines are tangent at opposite endpoints of the same diameter. Using Theorem 0., the two lines are perpendicular to the same line, so the are parallel. 3. is in the interior of and 5. Theorem 5.6, ###$ bisects. 33. For an point outside of a circle, there is never onl one or more than two tangents to the circle that passes through the point. There will alwas be two tangents. 34. Statements. 5 5, 5 8. Given. radius r 5 P 5 PE 5 PF 3. 5, 5 F F, 5 E E Reasons. ef. of radius Given 3. Postulate, Segment ddition 5. 5 F F 5. Substitution for and 6. E 5, E 5 F, 6. Theorem 0. 5 F 7. E 5 E 7. Substitution for, F, and F 8. E 5 E 8. Subtract 9. E and P are on the angle 9. Theorem 5.6 bisector of. 0. E 5 P PE 0. Postulate, Segment ddition Given. E E 5 8. Substitution for 3. E E Substitution for E 4. E ivide b. 5. PE E 5. Theorem 0. opright b Mcougal Littell, a division of Houghton Mifflin ompan. 3

3 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan. Statements Reasons 6. 5 F F 6. Statement F 4 7. Substitution for and F 8. F Subtract 4 9. (E) (E) 5 () 9. Pthagorean Theorem 0. 4 (E) 5 0. Substitution for E and. E 5 8 Ï. Solve for E.. PF F. Theorem (PF) (F) 5 (P) 3. Pthagorean Theorem 4. (PF) (F) 5 (E PE) 4. Substitution for P 5. r 8 5 (8 Ï r) 5. Substitution for PF, F, E, and PE 6. r 5 Ï 6. Solve for radius, r. Problem Solving 35. The wheel has radial spokes because each spoke has endpoints that are the center and a point on the circle. 36. The wheel has tangential spokes because each spoke intersects the circle in eactl one point. 37. E 5 E (, ) , ,098, ,46 ø 5 ø 4,46 miles 38. m R 5 m S 5 908, so R > S. lso, R > S because vertical angles are congruent. Therefore, nr, ns b the Similarit Postulate. So, 5 R because the ratio of corresponding S sides is the same. 39. a. ecause R is eterior to (Q and P is on (Q, QR > QP. b. ecause QR is perpendicular to line m it must be the shortest distance from Q to line m. Thus, QR < QP. c. It was assumed QP was not perpendicular to line m but QR was perpendicular to line m. Since R is outside of (Q ou know that QR > QP but part (b) tells ou that QR < QP which is a contradiction. Therefore, line m is perpendicular to QP. 40. ssume line m is not tangent to (Q. So, there is another point X on line m that is also on (Q. X is on (Q, so QX 5 QP. ut the perpendicular segment from Q to line m is the shortest such segment, so QX > QP. QX cannot be both equal to and greater than QP. The assumption that point X eists must be false. Therefore, line m is tangent to (Q. 4. Statements. SR and ST are tangent to (P.. Given Reasons. SR RP, ST TP. Tangent and radius are perpendicular. 3. RP 5 TP 3. ef. of circle 4. RP > TP 4. ef. of congruence 5. PS > PS 5. Refleive Propert 6. nprs > npts 6. HL ongruence Theorem 7. SR > ST 7. orresponding parts of congruent triangles are congruent. 4. a. The slope of the line perpendicular to line l through is 3 4, so the slope of line l is 4 3. b. 5 m b (4) b b b The equation for l is c. 3 (4) 5 r 5 5 r 5 5 r d. The -intercept is 5 and the radius is 5. So, the 3 distance from the -intercept to ( is Mied Review 43. m m 5 m 58 m m ( 3) (4 7)

4 hapter 0, continued > > > 5 > So, the third side must be greater than 5 and less than. 4 > 4 > > 7 5 > So, the third side must be greater than 7 and less than 5. Lesson Guided Practice (pp ). TQ is a minor arc. m TQ QRT is a major arc. m QRT 5 m QR m RT TQR is a semicircle. m TQR QS is a minor arc. m QS 5 m 5. TS is a minor arc. m TS RST is a semicircle. m RST m 5 m. 8. MN and QR m RS > because the are in congruent circles and PQ have the same measure but the are not congruent because the are arcs of circles that are not congruent. 0. Eercises (pp ) Skill Practice. If and E are congruent central angles of (, then and E are congruent.. You need to know that the radii of two circles are the same in order to show that the two circles are congruent. 3. is a minor arc. m is a minor arc. m m E m is a minor arc. m 5 m m E is a minor arc. m E 5 m is a minor arc. m 5 m E m E is a semicircle. m is a major arc. m 5 m m E is a major arc. m E 5 m E m ; QRS is a semicircle because ou know that QS is a diameter.. m > because the are in the same circle and m 5 m. 3. LP and MN have the same measure but the are not congruent because the are arcs of circles that are not congruent. 4. VW > XY because the are in congruent circles and m VW 5 m XY. 5. The statement is incorrect because ou can tell that the circles are congruent. The circles have the same radius,. 6. P 0 m m m m ; np is a right triangle. P P Ï 5 8. G E 0 00 F m GE If m GH 5 508, point H must be 08 beond point E, placing it on EF. Or, point H is 308 beond point F, placing it on EF. 9. R E 0 m E 5 m m m m E m E (m m m m E ) npq is a right triangle with m PQ and m QP npq is a right triangle with m PQ and m QP So, m Q 5 m QP m QP Therefore, m U opright b Mcougal Littell, a division of Houghton Mifflin ompan. 34

5 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan.. a. tan X m X ø m ø b. tan Y m Y ø 53.8 m ø 53.8 c. m 5 m m Problem Solving. The measure of the arc is The measure of each arc in the outermost circle of the dartboard is a The measure of the arc surveed b the camera is 708. b per minute 5 7 minutes It takes the camera 7 minutes to surve the area once. c. The camera must go counterclockwise and another 858 clockwise to return to the same position. So, per minute 5 37 minutes. It will take the camera 37 minutes. d. In 5 minutes, the camera can go 08(5) The camera will go 508 counterclockwise and then 008 clockwise. So, the camera will be 008 from wall after 5 minutes. 5. a per minute 60 min fter 0 minutes, the minute hand will move 0(68) per hour h The hour hand moves 308 per hour. 308 (0 min) 5 08 in 0 minutes 60 min t :0, the hour hand is from the. The minute hand is 08 from. The minor arc between the hour and minute hand is b. You want the arc between the hour and minute hand to be 808. Use guess, check, and revise. t :40, the minute hand is 40(68) from. The hour hand 308 moves (40 min) 5 08 in 40 minutes. The hour 60 min hand is from. The arc is t :38, the minute hand is 38(68) 5 88 from. The hour hand moves 308 (38 min) 5 98 in 38 minutes. The hour hand is 60 min from. The arc is Mied Review ( ) The slopes of the lines are different, so the are not parallel Y The slopes are the same, so the lines are parallel. Z9 X9 P ( )( 3) X ( 5)( 7) Z ( 6)( 6) (z 3) 5 (z 3)(z 3) 5 z 3z 3z 9 5 z 6z (3 7)(5 4) (z )(z 4) 5 z 4z z 4 5 z 5z 4 Lesson Guided Practice (pp ). m m m m m 5 08 m (80 ) m (88) (80 ) m E 5 (80 )8 5 (80 8) m E 5 m m E QR 5 ST QU 5 QR 5 (3) 5 6 Y 35

6 hapter 0, continued 8. Q 5 QU U Q 5 6 Q Q Eercises (pp ) Skill Practice. Sample answer: Point Y bisects XZ if XY > YZ.. If two chords of a circle are perpendicular and congruent, one of them does not have to be a diameter. square inscribed in a circle is an eample of two chords that are congruent and perpendicular. 3. m 5 m E m m m m 5 38 m EG 5 EJ Theorem 0.5, bisects Theorem 0.5, LN bisects PM ecause SQ and TQ are radii, the have the same measure and are equidistant from the center, so b Theorem 0.6, the are congruent and are equidistant from the center, so b Theorem 0.6, the are congruent. Theorem 0.5, 5 (3 ) EF and HG are congruent, so b Theorem 0.6, the are equidistant from Q ecause is a perpendicular bisector of, it is a diameter b Theorem ecause JH is a diameter and it is perpendicular to FG, it bisects FG and FG b Theorem ecause NP and LM are equidistant from the center, NP > LM b Theorem ; hoice is true because the are radii of the circle. Theorem 0.5 choice is true. hoice is true because the two triangles are congruent b the SS ongruence Postulate. 6. From the diagram, 5 and EF 5 4. However, b Theorem 0.6, 5 EF. 7. You do not know that, therefore ou cannot show that >. 8. is a perpendicular bisector of, so b Theorem 0.4, it is a diameter. 9. The two triangles are congruent using the SS ongruence Postulate. So, is the perpendicular bisector of. Theorem 0.4, is a diameter. 0. the Pthagorean Theorem, E 5 4. E Þ E, so is not a perpendicular bisector of. Therefore, is not a diameter.. Using the facts that np is equilateral which makes it equiangular and that m 5 308, ou can conclude that m P 5 m P You now know that m which makes >. np > np using the SS ongruence Postulate because P > P and P > P. Using corresponding parts of congruent triangles are congruent, >. long with >, ou have n > n using the SSS ongruence Postulate.. a. The converse of Theorem 0.5 is: If a diameter of a circle bisects a chord and its arc, then the diameter is perpendicular to the chord. The converse is different from Theorem 0.4 in that in the hpothesis, it is known that the first chord is a diameter. b. P S R T Statements. PT > TR, PS > SR Q. Given Reasons. T > T. Refleive Propert 3. P > R 3. ll radii in a circle are congruent. 4. npt > nrt 4. SSS ongruence Postulate 5. PT > TR 5. ongruent parts of congruent ns are congruent. 6. m TP m TR 6. Linear Pair Postulate m TP Substitution Propert 8. T PR 8. ef of opright b Mcougal Littell, a division of Houghton Mifflin ompan. 36

7 hapter 0, continued c. Using the converse of Theorem 0.5, ou know that PT > RT and QT PR. the Refleive Propert, QT > QT. Using the SS ongruence Postulate, npqt > nrqt. PQ > RQ because corresponding parts of congruent triangles are congruent. So, PQ 5 RQ. 3. From the diagram, m 5 8. Theorem 0.3, m 5 m. Let 8 5 m 5 m. Then m m m is an even number because (360 ) is even. P 8. ecause >, P > P b the definition of congruent arcs. P, P, P, and P are all radii of (P, so P > P > P > P. Then np > np b the SS ongruence Postulate, so corresponding sides and are congruent. 9. a. The longer chord is closer to the center. b. The length of a chord in a circle increases as the distance from the center of the circle to the chord decreases. c. Let c be the radius. d Let <. b opright b Mcougal Littell, a division of Houghton Mifflin ompan. T R In npt, P 5 0 and T 5 8. m PT 5 sin In npr, P 5 0 and R 5 6. m PR 5 sin m 5 m R m R Problem Solving 5. In order for >, should be congruent to. 6. To find the center of the cross section, ou can () construct the perpendicular bisector of the control panel, etend this segment to the other control panel, and then find the midpoint of the segment. Or ou can () draw two diagonals from the top of one control panel to the bottom of the other. The center will be at their intersection. 7. Statements. >. Given Reasons. P, P, P, and P. Given are radii of (P. 3. P > P > P > P 3. ll radii of same circle are congruent. 4. np > np 4. SSS ongruence Postulate 5. P > P 5. orr. parts of > ns are >. 6. m P 5 m P 6. ef. of > 7. P and P 7. Given are central s. 8. m 5 m 8. ef. of arc measure 9. > 9. ef. of > the Pthagorean Theorem, c 5 b and c 5 d So, b 5 d. Since <, b > d. Therefore, the length of a chord in a circle increases as the distance from the center decreases. 30. a. r 5 60 (r 80) r 5 5,600 r 60r r 5 3,000 r 5 00 The radius of the circle is 00 feet. b. S Ï fr Ï (0.7)(00) ø The car s speed is about 45.7 miles per hour. 3. Given QS is the perpendicular bisector of RT in (L. Suppose L is not on QS. Since LT and LR are radii of the circle, the are congruent. With PL > PL b the Refleive Propert, ou now have nrlp > ntlp using the SSS ongruence Postulate. orresponding angles RPL and TPL are now congruent and the form a linear pair. This makes them right angles and leads to QL being perpendicular to RT. Using the Perpendicular Postulate, L must be on QS and thus QS must be a diameter. 3. raw radii L and LF. So, L > LF, L > L (Refleive Propert) and since EG F, nl > nlf b the HL ongruence Theorem. Then, corresponding sides and F are congruent, as are corresponding angles L and FL. the definition of congruent arcs, G > FG. 33. ase : Given: EF, EG, EG EF Prove: > raw radii E and E. E > E and EF > EG. lso, since EF and EG, nef and neg are right triangles and are congruent b the HL ongruence Theorem. orresponding sides F and G are congruent, so F 5 G and, b the Multiplication Propert of Equalit, F 5 G. Theorem 0.5, EF bisects and EG bisects, so 5 F and 5 G. Then b the Substitution Propert, 5 or >. 37

8 hapter 0, continued ase : Given: EF, EG, > Prove: EF > EG raw radii E and E, E > E. Theorem 0.5, EF bisects and EG bisects, so 5 F and 5 G. We know 5, so b the Substitution Propert, F 5 G. the ivision Propert of Equalit, F 5 G, or F > G. nef and neg are right triangles and are congruent b the HL ongruence Theorem. It follows that corresponding sides EF and EG are congruent. 34. Q T P The point where the tire touches the ground is a point of tangenc. Theorem 0., line g is perpendicular to radius of the circle, TP. ecause i g, PT. Theorem 0.5, PQ bisects and. Mied Review ( 0)8 ( 0) J K Þ 69 is not tangent to ( at because m Þ 908 and radius is not perpendicular to. 3. m EFG 5 m EF m FG m FG m FG m EG m EFG m 5 m m m 5 m m m 5 m m m Lesson 0.4 Investigating ctivit 0.4 (p. 67). nswers will var.. nswers will var. 3. The measure of an inscribed angle is one half the measure of the corresponding central angle. 37. J M L JKLM is a rectangle. M K JKLM is a rhombus. L Quiz (p. 670) is tangent to ( at because m and radius is perpendicular to. 0.4 Guided Practice (pp ). m HGF 5 m HF 5 (908) m TV 5 m TUV 5 (388) m ZXW 5 m ZYW To frame the front and left side of the statue in our picture, make the diameter of our circle the diagonal of the rectangular base. This diagonal connects the upper left corner to the bottom right corner. 5. m m m m m S m U c8 (c 6) c 5 86 c 5 6 m T m V opright b Mcougal Littell, a division of Houghton Mifflin ompan. 38

9 opright b Mcougal Littell, a division of Houghton Mifflin ompan. hapter 0, continued 0.4 Eercises (pp ) Skill Practice. If a circle is circumscribed about a polgon, then the polgon is inscribed in the circle.. The diagonals of a rectangle create two right triangles. Theorem 0.9 tells ou the hpotenuse of each of these triangles is a diameter of the circle. 3. m 5 m 5 (848) m F m G 5 m F 5 (708) m LM m N 5 m LM 5 (08) m RS 5 m Q 5 (678) m TV 5 m U 5 (308) m UV m TV m YW 5 m X 5 (758) m WX m XY m YW From the diagram, the measure of RS is 908. So, the measures of the arcs add up to You can either change the measure of Q to 408 or change the measure of QS to > because the intercept the same arc,. > because the intercept the same arc,.. JMK > KLJ because the intercept the same arc, JK. MKL > LJM because the intercept the same arc, LM.. WXZ > ZYW because the intercept the same arc, ZW. XWY > YZX because the intercept the same arc, XY. 3. m R m T m S m Q m m F k k 5 0 k 5 60 m E m G m m m J 5 m KLM 5 (308 08) 5 08 m J m L a a 5 60 a 5 0 m K 5 m JML 5 ( ) 5 98 m K m M b b 5 88 b 5 6. ; m m 5 m 5 (608) a. There are 5 congruent arcs The measure of each inscribed angle is the measure of the arc. (78) There are 5 inscribed angles. Sum 5 5(368) b. There are 7 congruent arcs ø The measure of each inscribed angle is the measure of the arc. (5.48) ø There are 7 inscribed angles. Sum ø 7(5.78) ø 808. c. There are 9 congruent arcs The measure of each inscribed angle is the measure of the arc. (408) There are 9 inscribed angles. Sum 5 9(08) ; m EFG 5 m EG (8 0)8 5 ( 40) In a parallelogram, the opposite angles are congruent. In an inscribed parallelogram, opposite angles are supplementar. rectangle is a parallelogram with congruent supplementar opposite angles. The m R is square can alwas be inscribed in a circle because its opposite angles are 908 and thus are supplementar.. rectangle can alwas be inscribed in a circle because its opposite angles are 908 and thus are supplementar.. parallelogram cannot alwas be inscribed in a circle because its opposite angles are not alwas supplementar. 3. kite cannot alwas be inscribed in a circle because its opposite angles are not alwas supplementar. 4. rhombus cannot alwas be inscribed in a circle because its opposite angles are not alwas supplementar. 5. n isosceles trapezoid can alwas be inscribed in a circle because its opposite angles are supplementar. 39

10 hapter 0, continued 6. 3 J 5 K 4 JK is a diameter. The altitude from to is also a diameter. n is a right triangle, so sin Using the smaller triangle, sin 5 alt 3. So, alt 3 alt 5 5. Therefore, the diameter JK is also 5. Problem Solving 7. 0,000 km 00,000 km 5 00,000 0,000 00, ,000 Moon is 0,000 km from moon. 8. Place the carpenter s square so the endpoints of the square and the verte of the square are on the circumference of the circle, then connect the endpoints. 9. The hpotenuse of the right triangle inscribed in the circle is the diameter of the circle. So, double the length of the radius to find the length of the hpotenuse. 30. the rc ddition Postulate, m EFG m GE and m FG m EF Using the Measure of an Inscribed ngle Theorem, m EG 5 m F, m EFG 5 m, m EG 5 m G, and m FG 5 m E. the Substitution Propert, m m F , so m m F Similarl, m E m G Let m 5 8. ecause Q and Q are both radii of (Q, Q > Q and nq is isosceles. ecause and are base angles of an isosceles triangle, >. So, b substitution, m 5 8. the Eterior ngles Theorem, m Q 5 m m 5 8. So, b the definition of the measure of a minor arc, m 5 8. ivide each side b to show that 8 5 m. Then, b substitution, m 5 m. 3. Given: is inscribed in (Q. Point Q is in the interior of. Prove: m 5 m Plan for Proof: onstruct the diameter of (Q and show m 5 m and m 5. Use the rc ddition Postulate and the ngle ddition Postulate to show m 5 m m. 33. Given: is inscribed in (Q. Point Q is in the eterior of. Prove: m 5 m Plan for Proof: onstruct the diameter of (Q and show m 5 m and m 5 m. Use the rc ddition Postulate and the ngle ddition Postulate to show m 5 m m. 34. Given: and are inscribed angles. Prove: > O Paragraph Proof: Theorem 0.7, m 5 m and m 5 m. substitution, m 5 m. the definition of congruence, >. 35. ase : Given: (T with inscribed n. is a diameter of (T. Prove: n is a right triangle. T Plan for Proof: Use the rc ddition Postulate to show E that m E 5 m and thus m Then use the Measure of an Inscribed ngle Theorem to show m 5 908, so that is a right angle and n is a right triangle. ase : Given: (T with inscribed n. is a right angle. Prove: is a diameter of (T. Plan for Proof: Use the Measure of an Inscribed ngle Theorem to show the inscribed right angle intercepts an arc with measure (908) Since intercepts an arc that is half of the measure of the circle, it must be a diameter. 36. In the figure, n is a right triangle with being the right angle. Using Theorem 0., is perpendicular to radius, it is tangent to ( at point. 37. HJ GJ 5 GJ ; in a right triangle, the altitude from the right FJ angle to the hpotenuse divides the hpotenuse into two segments. The length of the altitude is the geometric mean of the lengths of these two segments. 38. FJ 5 6 in., JH 5 in. HJ GJ 5 GJ FJ Ï 3 GJ 5 Ï 3 in. GK 5 GJ 5 ( Ï 3 ) 5 4 Ï 3 in. opright b Mcougal Littell, a division of Houghton Mifflin ompan. 330

11 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan. 39. Show that m P is less than or equal to 458. The diagram shown represents of the fuel booster design. is equal to of 3608 or about P bisects and P, so m ø P is also the perpendicular bisector of, and contains point. n and np are right triangles with in both triangles..5.5 sin ø cos P ø m P 5 cos ø 688 Since np is isosceles, the base angles are equal. m P 5 m P m P m P m P So m P is less than 458. Mied Review ø 8.4 ø ø E 9 G9 H 9 F9 F G H9 E Q P S9 R Q9 S 9 R Lesson Guided Practice (pp ). m 5 (08) m RST 5 m T 5 (988) m XY 5 m X 5 (808) (958 8) m FJG 5 (m FG m KH ) (a8 448) 60 5 a a 6. sin TQS m TQS ø m TQR 5 (m TQS) ø (36.878) ø m TQR 5 (m TUR m TR ) (8 (360 8)) ø 0.5 Eercises (pp ) Skill Practice. The points,,, and are on a circle at P. If m P 5 (m m ) then P is outside the circle.. If m 5 08, then the two chords intersect on the circle. Theorem 0., m 5 (m m ) 5 (m 08) 5 m. This is consistent with the measure of an Inscribed ngle Theorem (Lesson 0.4). 3. m 5 m 4. m 5 m EF m 78 5 m EF m m EF 5. m 5 (608) ; ecause is not a diameter, m Þ 808. m 5 m Þ (808) Þ

12 hapter 0, continued ( ) 5 (308) (8 458) (80 )8 5 (308 ( 30)8) (80 ) (478 38) 5 (34) (48 8) [(3 )8 ( 6)8] 68 5 (3 ) ( 6) ; m 4 5 (808 08) 5 (008) The error is that the given measurements impl two different measures for E. Using Theorem 0., (m E 608) m E m E Using Theorem 0.3, 58 5 (608 m E ) m E m E If ###$ PL is perpendicular to KJ at K, then m LPJ 5 908, otherwise it would measure less than 908. So, m LPJ a. (08) (8 408) b. c8 5 (b8 a8) c 5 b a 7. m Q 5 (m EGF m EF ) [(360 )8 8] So, m EF m R 5 (m GEF m FG ) [(360 )8 8] So, m FG m GE m EF m FG So, m GF m 5 (m m ) (78 38) m (0) m (0) m m m a. t t b. For the diagram on the left, m 5 m m 5 m For the diagram on the right, m 5 (3608 m ) m m m m m 5 (808 m ) opright b Mcougal Littell, a division of Houghton Mifflin ompan. 33

13 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan. c. m 5 (808 m ) m m m m So, these equations give the same value for m when is perpendicular to t at point. 0. Let 8 5 m XW 5 m ZY. Then m WZ 5 (00 )8 and m XY 5 (3608 (00 )8) 5 (60 )8 m P 5 (m WZ m XY ) 5 [(00 )8 (60 )8] 5 (40) m H (m m E ) m H (858 m E ) m E m E m F 5 m E m EF m J 5 (m m F ) (m 58) m m m FGH 5 (m m F ) (858 (08 m E )) m E m E Problem Solving. m m 5 ( ) 5 (508) 5 58 m 5 (808) ( ) 5 (008) m 5 (808 8) (808 8) amera will have a 308 view of the stage when the arc measuring 308 is reduced to an arc measuring 08. You should move the camera closer to the stage E 4000 mi 400. mi Not drawn to scale sin m ø m ø (88.68) ø 77.8 Let m 5 8. m 5 (m E m ) 77.8 ø [(3608 8) 8] ø.8 The measure of the arc from which ou can see is about [(80 )8 8] (80 ) (80 )8 5 (80 76) The measures of the arcs between the ground and the cart are 768 and ase : Given: intersects chord at point on (Q. contains the center of (Q. Prove: m 5 m Q ase Paragraph proof: Theorem the definition of perpendicular, m ecause So, m 5 is a diameter, m m. ase : Given: intersects chord at point on (Q. The center of the circle, Q, is in the interior of. Prove: m 5 m P Q P ase 333

14 hapter 0, continued Plan for proof: raw diameter P. Use the ngle ddition Postulate and Theorem 0.7 to show that m m P. Use the rc ddition Postulate to show that m P m P. ase 3: Given: intersects chord at point on (Q. The center of the circle, Q, is in the eterior of. Prove: m 5 m P Q ase 3 Plan for Proof: raw diameter P. Use the ngle ddition Postulate and Theorem 0.7 to show that m m P. Use the rc ddition Postulate to show that m m P. 8. Given: hords and intersect. Prove: m 5 (m m ) Statements. hords and intersect.. Given. raw.. Given Reasons Then, m 5 (m m ). P Q 4 3 R ase ase : raw PR. Use the Eterior ngle Theorem to show that m 3 5 m m 4, so that m 5 m 3 m 4. Then use Theorem 0. to show that m 3 5 PQR and m 4 5 PR. Then, m m 5 (m PQR m PR ). W 3 Z 4 X Y m ase 3 ase 3: raw XZ. Use the Eterior ngle Theorem to show that m 4 5 m 3 m WXZ, so that m 3 5 m 4 m WXZ. Then use the Measure of an Inscribed ngle Theorem to show that m 4 5 m XY and m WXZ 5 m WZ. Then, m 3 5 (m XY m WZ ). 30. Q P T 3. m 5 m 3. Eterior ngle Theorem m 4. m 5 m 4. Theorem m 5 m 5. Theorem m 5 m m 6. Substitution Propert 7. m 5 (m m ) 9. ase 7. istributive Propert ase : raw. Use the Eterior ngle Theorem to show that m 5 m m, so that m 5 m m. Then use Theorem 0. to show that m 5 m and the Measure of an Inscribed ngle Theorem to show that m 5 m. R Given: PQ and PR are tangents to a circle. Prove: QR is not a diameter. Paragraph Proof: ssume QR is a diameter. Then m QTR 5 m QR Theorem 0.3, m P 5 (m ( ) 5 08 The m P cannot equal 08, so QR cannot be a diameter cm E 5 3, E 5 5 E 5 E 3 cm 3 5 5, E 5 cm m E 5 sin 3 ø 8.48 m 5 m E m E ø 48% 3608 bout 48% of the circumference of the bottom pulle is not touching the rope. opright b Mcougal Littell, a division of Houghton Mifflin ompan. 334

15 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan. Mied Review 3. ecause P9 P , the scale factor is K 5 3. This is 4 a reduction. 33. ecause P9 P , the scale factor is K 5 5. This is 3 an enlargement Ï 7 4()(6) () Ï or () 6 Ï () 4()() () 5 6 Ï or (8) 6 Ï (8) 4()(6) () Ï Ï 6 4()(0) () Ï 76 ø.36 or ø Ï 5 4()(9) () Ï 97 ø. or ø Ï 3 4(4)() (4) Ï 85 ø.33 or ø.08 8 Quiz (p. 686). m m m m m 5 m 5 (958) So, z m E m G m F m H m F 5 80 m F m GHE 5 m F 5 (908) So, z m K m M m L m J ( ) ((7) ) m JKL 5 m M 5 (38) 5 68 So, z ( ) 5 (908) (748 8) 5 (58) (8 878) E 4000 mi mi Not drawn to scale 4000 sin 5 m ø m ø (88.58) ø 778 Let m 5 8. m 5 (m E m ) 778 ø [(3608 8) 8] 354 ø ø ø 3 The measure of the arc from which ou can see is about

16 hapter 0, continued Mied Review of Problem Solving (p. 687). a. r 3 5 (r ) r 9 5 r 4r r r.5 5 r The radius of the discus circle is.5 meters. b The official is 3.5 meters form the center of the discus circle.. m XYZ m XQZ m YZ 5 m XYZ 5 (68) a b (3) The distance between the lowest point reached b the blades and the ground is 99 feet. c cos ø ø (3.45) ø 6.9 The distance from the tip of one blade to the tip of another is about 6.9 feet. 4. a b ; 8 angles and 9 total spokes. There are 7 spokes between the two spokes. c. The wheel itself is feet tall. The diameter is the total height, or 40 feet. The radius is half the height, or 70 feet. 5. nswers will var. 6. a. 8 5 (m NM m LK ) 5 ( ) b. m LK 5 m MN m LM m MN m KN 5 m LM (m m ) 8 5 m m m 5 8 m 8 5 (m m ) 8 5 m m m 5 8 m 8 m 5 8 m Lesson 0.6 m m Investigating ctivit 0.6 (p. 688). The products E p E and E p E are the same.. The products E p E and E p E are the same. 3. E p E 5 E p E 4. PT p QT 5 RT p ST 9 p p ST ST 3 5 ST 0.6 Guided Practice (pp ). 6 p (6 9) 5 5 p (5 ). 3 p 5 6 p p [3 ( )] 5 ( ) p [( ) ( )] 3( 5) 5 ( )() ( 5)( 3) is etraneous (3 ) ( 5 is etraneous.) p (5 ) p ( 0) ( 8)( 8) ( 5 8 is etraneous.) opright b Mcougal Littell, a division of Houghton Mifflin ompan. 336

17 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan. 7. Use Theorem p ( 4) Ï 4 4()(5) () Ï Ï Ï 74 ( 5 7 Ï 74 is etraneous.) 8. Use Theorem 0.4. p p Use Theorem p (8 ) 5 p ( 9) ( 6)( 45) ( 5 45 is etraneous.) 0. E cannot be equal to E because that would make E tangent to the circle. E must be less than E.. It is appropriate to use the approimation smbol in the solution to Eample 4 because it is stated in the problem that each moon has a nearl circular orbit. 0.6 Eercises (pp ) Skill Practice. The part of the secant segment that is outside the circle is called an eternal segment.. tangent segment intersects the circle in onl one point while the secant segment intersects the circle in two points. 3. p 5 0 p p ( 3) 5 0 p p ( 8) 5 6 p ( 4)( ) ( 5 is etraneous.) 6. 8 p (8 ) 5 6 p (6 0) p ( 4) 5 5 p (5 7) ( 6)( 0) ( 5 0 is etraneous.) 8. ( ) p [( ) ( 4)] 5 4 p (4 5) ( )( ) ( 5)( 4) ( 5 4 is etraneous.) p (9 7) ( 5 is etraneous.) p ( ) ( 4) 5 p ( ) The error is that the wrong segment lengths are being multiplied. Use Theorem 0.5. F p F 5 F p F 4 p (4 ) 5 3 p (3 5) p ( 3) 5 p 4. p ( Ï 3 ) 5 p ( ) ( )( 3) ( 5 3 is etraneous.) 6. ; p 5 p ( 6) ( 6)( ) ( 5 is etraneous.) 7. MN 5 PN p (PN PQ) 5 6 p (6 PQ) PQ PQ 8 5 PQ 337

18 hapter 0, continued 8. RQ 5 RS p (RS SP) RQ 5 4 p (4 ) RQ PQ RQ 5 RP PQ PQ 5 3 PQ ø p ( E) 5 8 p (8 6) E r P p E 5 (r P) p (r P) 4 p 6 5 (r 4) p (r 4) 4 5 r r Ï 0 5 r Problem Solving 0. p 5 6 p (50 6) 5,656 ø The distance from the end of the passage to either side of the mound is about 08 feet.. Given: and are chords that intersect at E. Prove: E p E 5 E p E Statements E Reasons. and are chords. Given that intersect at E.. raw and.. Two points determine a line. 3. > 3. and intercept the same arc. 4. > 4. and intercept the same arc. 5. ne, ne 5. Similarit Postulate 6. E E 5 E E 6. orresponding sides are proportional. 7. E p E 5 E p E 7. ross Product Propert. p ( E) 5 p ( ) () ()(E) 5 () ()() () ()(E) () 5 ()() () ()(E) () 5 3. Given: E is a tangent segment and E is T a secant segment E passing through the center. Prove: (E) 5 E p E Proof: raw radius T. Theorem 0., E T. the Pthagorean Theorem, E T 5 ET. the Segment ddition Postulate, ET 5 E T. So, E T 5 (E T) b substitution. Simplif the equation. E T 5 (E T) E T 5 E (E)(T) T T and T are both radii, so the are equal. Substitute T for T. E T 5 E (E)(T) T E 5 E(E T) E 5 E(E ) E 5 E p E 4. 4 p N 5 6 p 8 N 5 The length N is centimeters. Sparkles travel from 6 cm to in 5 3 seconds. So, sparkles must travel cm/sec from to N in 3 seconds. The sparkles must travel at a cm rate of 3 sec 5 4 cm/sec. 5. Given: E and E are secant segments. Prove: E p E 5 E p E E Paragraph proof: raw and. and intercept the same arc, so >. E > E b the Refleive Propert of ongruence, so ne, ne b the Similarit Theorem. Then, since lengths of corresponding sides of similar triangles are proportional, E E 5 E. the ross Product Propert, E E p E 5 E p E. opright b Mcougal Littell, a division of Houghton Mifflin ompan. 338

19 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan. 6. Given: E is a tangent segment. E is a secant segment. Prove: E 5 E p E E Paragraph proof: raw and. m 5 m and m E 5 m. So, > E. E > E b the Refleive Propert of ongruence, so ne, ne b the Similarit Theorem. Then, since lengths of corresponding sides of similar triangles are proportional, E E 5 E. the ross Product Propert, E (E) 5 E p E. 7. a. m 5 m 5 (08) b. m 5 m EF 5 608, so > EF. lso, > FE b the Vertical ngles Theorem. So, n, nfe b the Similarit Theorem. c. EF 5 F d. Use Theorem 0.6. EF 5 F p F 5 p ( 0 6) 5 ( 6) e. 5 ( 6) 0 5 ( 6) (3 50)( ) or Length cannot be negative, so f. EF , so the ratio of nfe to n is to. So, F 5. Let E 5 and 5. Theorem 0.4, 8. N E p 5 p p 5 6 p Ï 30 Length cannot be negative, so 5 Ï 30. E 5 5 Ï 30 O P 5 The distance from the origin to P9 is 5 (Pthagorean Theorem). ON is the diameter, so its length is. (NP9) 5 (ON) (OP9) P9 NP9 5 Ï 9. Theorem 0.6, (OP9) 5 PP9 p NP9 5 5 PP9 p Ï 9 5 Ï 9 5 PP9 5 Ï 9 5 PP9 9 The distance d is 5 Ï 9 units. 9 Mied Review 9. Ï (0) 8 5 Ï Ï Ï 5 (4) (6 ) 5 Ï Ï Ï Ï [ (6)] (3 6) 5 Ï 4 (3) P R 50 5 E F Q 5 Ï Ï cos QR QR 5 8 cos 408 ø 0.4 tan PR PR 5 8 tan 508 ø 6.7 F 5 E EF F 5 Ï m 5 m F m 5 m F m m m m is a semicircle, so m E is a semicircle, so m E does not bisect chord RS, so is not a diameter. 339

20 hapter 0, continued bisects and is perpendicular to, so is a diameter. 4. does not bisect chord, so is not a diameter. Problem Solving Workshop 0.6 (p. 696). RS RQ 5 RQ RT 4 RQ 5 RQ RQ 5 RQ 3 RQ 5 5 RQ 5 Ï 5 RQ 5 Ï 3. a. E m E 5 m m E m E m m 5 m E 5 (3608 m ) m So, m E 5 m or E >. lso > b the Refleive Propert of ongruence. So, ne, n b the Similarit Theorem. b. E E E E 56 5 E 3 5 E (5 ) ( w) 5 ( z) ( z) w 5 ( z) w Etension (pp ) in. P in. in. in. l k m. in. in. 5. The locus of points consists of two points on l each 3 centimeters awa from P. 3 cm 3 cm P 6. The locus of points consists of four points on a circle with center at Q and a radius of 5 centimeters. The four points are the intersections of the circle and two lines parallel to and 3 centimeters awa from m. m 3 cm 5 cm Q 3 cm 7. The locus of points consists of a semi-circle centered at R with radius 0 centimeters. The diameter bordering the semi-circle is 0 centimeters from k and parallel to k. 0 cm 0 cm R 8. The portions of line l and m that are no more than 8 centimeters from point P and the portion of the circle, including its interior, with center P and radius 8 centimeters that is between lines l and m. 5 cm 8 cm P 8 cm 5 cm l k m l l k m opright b Mcougal Littell, a division of Houghton Mifflin ompan. 340

21 hapter 0, continued 9. ft 6 ft ft dog 3 ft house 5 ft 4 ft 9 ft 7. Three seismographs are needed to locate an earthquake s epicenter because two circles intersect in two points. You would not know which point is the epicenter. You need the third circle to determine which point is the epicenter. 0.7 Eercises (pp ) opright b Mcougal Littell, a division of Houghton Mifflin ompan. Lesson 0.7 dog 0.7 Guided Practice (pp ). 5 r ( h) ( k) 5 r ( ()) ( 5) 5 7 ( ) ( 5) r 5 Ï (3 ) (4 4) 5 Ï 0 5 ( h) ( k) 5 r ( ) ( 4) 5 ( ) ( 4) r 5 Ï ( ) ( 6) 5 Ï (3) (4) 5 5 ( h) ( k) 5 r ( ) ( 6) 5 5 ( ) ( 6) ( 4) ( 3) 5 6 ( 4) ( (3)) 5 4 The center is (4, 3) and the radius is 4. (4, 3) 6. ( 8) ( 5) 5 ( (8)) ( (5)) 5 The center is at (8, 5) and the radius is. (8, 5) 3 6 Skill Practice. The standard equation of a circle can be written for an circle with known center and radius.. The location of the center and one point on a circle is enough information to draw the rest of the circle because the distance from the center to the known point is the radius of the circle. 3. The radius is The center is (, 3). The radius is. ( ) ( 3) The radius is The center is (5, 0). The radius is 0. ( 5) ( 0) 5 0 ( 5) The center is (50, 50). The radius is 0. ( 50) ( 50) 5 0 ( 50) ( 50) The center is (3,3). The radius is 9. ( (3)) ( (3)) 5 9 ( 3) ( 3) ( (4)) ( ) 5 ( 4) ( ) 5. ( 7) ( (6)) 5 8 ( 7) ( 6) ( 4) ( ) 5 5 ( 4) ( ) ( 3) ( (5)) 5 7 ( 3) ( 5) ( (3)) ( 4) 5 5 ( 3) ( 4) If (h, k) is the center of a circle with radius r, the equation of the circle should be ( h) ( k) 5 r not ( h) ( k) 5 r. ( (3)) ( (5)) 5 3 ( 3) ( 5) ; r 5 6 so r 5 4. d 5 r 5 (4)

22 hapter 0, continued 7. r 5 Ï (0 0) (0 6) 5 Ï (6) r 5 Ï ( 4) ( ) 5 Ï (3) 5 3 ( ) ( ) 5 3 ( ) ( ) r 5 Ï (3 ) (5 8) 5 Ï (4) (3) 5 5 ( (3)) ( 5) 5 5 ( 3) ( 5) ( 3) 5 6 The center is (0, 0). The center is (3, 0). The radius is 7. The radius is 4. (0, 0). ( ) 5 36 The center is (0, ). The radius is 6. (0, ) 3. ( 4) ( ) 5 The center is (4, ). The radius is. (4, ) 4. ( 5) ( 3) 5 9 The center is (5, 3). The radius is 3. (5, 3) (3, 0) 5. ( ) ( 6) 5 5 The center is (, 6). The radius is 5. 8 (, 6) 6. ; ( ) ( 4) 5 5 (0 ) (5 4) Þ ( 6 9) 5 4 ( 3) 5 4 The equation is a circle ( 8 6) ( ) ( 4) ( ) 5 The equation is a circle ( 4 3 ) 5 6 ( 4 4) 5 7 ( ) 5 7 The equation is a circle ( 5 4) ( ) 5 77 ( ) 5 77 The equation is a circle. 3. ( 4) ( 3) 5 9 The center is (4, 3) (4) 6 3 Þ 6 The line does not contain the center of the circle. Solve for in both equations and set them equal to find points of intersection. ( 4) ( 3) 5 9 ( 3) 5 9 ( 4) 3 5 Ï 9 ( 4) 5 3 Ï 9 ( 4) Ï 9 ( 4) Ï 9 ( 4) (3 3) 5 9 ( 4) opright b Mcougal Littell, a division of Houghton Mifflin ompan. 34

23 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan. Use the quadratic equation to solve for Ï (3) 4(5)(8) (5) or ecause has two solutions, the line intersects the circle at two points. The line is a secant. 3. ( ) ( ) 5 6 The center is (, ) () 4 Þ 8 The line does not contain the center of the circle. Solve for in both equations and set them equal to find points of intersection. ( ) ( ) 5 6 ( ) 5 6 ( ) 5 Ï 6 ( ) 5 Ï 6 ( ) 4 5 Ï 6 ( ) 6 5 Ï 6 ( ) ( 6) 5 6 ( ) Use the quadratic equation to solve for Ï (0) 4(5)(4) (5) Ï 80 0 The square root of a negative number does not eist. There are no solutions, so the line does not intersect the circle. The line is not a secant or a tangent. 33. ( 5) ( ) 5 4 The center is (5, ) (5) 3 Þ The line does not contain the center of the circle. t 5 5, (5) The point (5, ) is in the interior of the circle, so the line is a secant. 34. ( 3) ( 6) 5 5 The center is (3, 6) (3) The line contains the center of the circle, so it is a secant that contains a diameter. 35. Let radius (O be k, then center (: (5k, 0), point: (63, 6). r 5 Ï (5k 63) (0 6) r 5 Ï 5k 890k 45 4k 5 Ï 5k 890k 45 6k 5 5k 890k k 890k k 5 Ï 4(09)(45) (09) k 5 k 5 5 or k k 5 5 because k is an integer. ( 5) 5 00 Problem Solving 36. a. Zone it enter Zone b. (3, 4) is in Zone. (6, 5) is in Zone 3. (, ) is in Zone. (0, 3) is in Zone. (, 6) is in Zone. Zone r 5 d 5 (4.8) r 5 d 5 (0.6) center: (3, 0) radius: 5 ( (3)) ( 0) 5 ( 3) 5 343

24 hapter 0, continued 39. center: (3, 0) radius: 5 7 ( 3) ( 0) 5 7 ( 3) nswers will var. 4. The height (or width) alwas remains the same as the figure is rolled on its edge. 4. a. Yes, cell towers X and Y cover part of the same Y 4 area, so this area ma Z receive calls from Tower X or Tower Y. 4 X b. Your home is within range of Tower Y, but our school is not in the range of an tower. So, ou can use our phone at home. c. it has better cell phone coverage because it is located entirel within the range of Tower Z. it is onl partiall covered b Tower X and Tower Y. 43. a. The center can be found at the intersection of the lines perpendicular to the tangents b (4) b b b (4) b b () The center is at (, 9). r 5 Ï ( 4) (9 5) 5 Ï (3) b. center: (, 9) radius: 5 ( ) ( 9) (, 9) 44. Given: circle passing through the points (, 0) and (, 0). Prove: The equation of the circle is k 5 with center at (0, k). onstruct the perpendicular bisector of chord with the endpoints (, 0) and (, 0). Using Theorem 0.4, this new chord is a diameter of the circle. Since the new chord is a segment of the -ais, the center of the circle is located at some point (0, k) which makes the equation of the circle ( k) 5 r or k 5 r k. Now consider the right triangle whose vertices are (0, 0), (0, k), and (, 0) with the distance from (0, k) to (, 0) being r, the radius of the circle. Using the Pthagorean Theorem, ou get k 5 r or r k 5. Substituting ou get k a. If r 5, there are two possible points of intersection. Use Theorem 0.. If the center of the circle is (8, 8), then the point of intersection for m and n is (0, 6). If the center of the circle is (0, 6), then the point of intersection is (8, 8). If r 5 0, these are also two possible points of intersection. ( has two possible centers. To find these centers, set the equation of the circle with radius 0 and center (8, 6) equal to the equation of the circle with radius 0 and center (0, 8). r m 5 0 center (8, 6) ( 8) ( 6) Ï 00 ( 8) r n 5 0 center (0, 8) ( 0) ( 8) Ï 00 ( 0) 6 Ï 00 ( 8) 5 8 Ï 00 ( 0) 00 ( 8) 5 ( Ï 00 ( 0) ) 00 ( 8) Ï 00 ( 0) 00 ( 0) ( 0) ( 8) Ï 00 ( 0) Ï 00 ( 0) ( 8) 5 00 ( 0) ( )( 6) or 5 6 opright b Mcougal Littell, a division of Houghton Mifflin ompan. 344

25 hapter 0, continued opright b Mcougal Littell, a division of Houghton Mifflin ompan. Possible centers for ( are (, 4) and (6, 0). If the center of ( is (6, 0), then line m is and line n is The point of intersection is 8 6 7, 7 7. If the center of ( is (, 4), then line m is and line n is The point of intersection is 9 7, The points of intersection and the center of ( are collinear. b. The locus of intersection points of m and n for all possible values of r for ( is the intersection of all circles centered (8, 6) and (0, 8). ( 8) ( 6) 5 ( 0) ( 8) (8, 6) m Mied Review (0, 8) r 5 r P 5 l w 5 () (9) in. 47. P 5 4s 5 4(8) 5 7 ft ø 40.6 P ø ø 37.6 m πr 5 (3.4)(7) cm πd 5 (3.4)(60) in πd 5 (3.4)(48) d n 5. 5 r 5 (9 r) 5 r 5 8 8r r r 8 5 r 53. r 5 (r 5) r 44 5 r 30r r 7. 5 r 54. r 8 5 (r 0) r r 40r r Quiz (p. 705). 6 p 5 8 p 9. 7 p (7 5) 5 6 p (6 ) ( ) 4. center: (, 4) radius: 6 7() ( ) ( 4) ( ) ( 4) center: (5, 7) 6. point: (5, 3) r 5 Ï (5 5) (7 (3)) 5 Ï (4) 5 4 ( 5) ( (7)) 5 4 ( 5) ( 7) center of tire: (4, 3) 4 radius of tire:. ( (4)) ( 3) 5. ( 4) ( 3) center of rim: (4, 3) radius of rim: 7 ( (4)) ( 3) 5 7 ( 4) ( 3) 5 49 Mied Review of Problem Solving (p. 706). a. center: (0, 30) radius: 0 ( 0) ( 30) 0 ( 0) ( 30)

26 hapter 0, continued b. t E(5, 5): ( 0) ( 30) 400 (5 0) (5 30)? (5)? You can receive the signal at E. t F(0, 0): ( 0) ( 30) 400 (0 0) (0 30)? 400 (0) (0)? > 400 You cannot receive the signal at F. t G(0, 6): ( 0) ( 30) 400 (0 0) (6 30)? 400 (4)? You can receive the signal at G. t H(35, 30): ( 0) ( 30) 400 (35 0) (30 30)? 400 5? You can receive the signal at H.. a. Y X Z The areas covered b the cell towers overlap, so there are areas that can transmit calls using more than one tower. b. Your house is located within range of tower X. You can use our cell phone at our house. You cannot use our cell phone at our friend s house. c. it has better coverage from the cell phone towers p p The track is not circular because the product of the lengths of one chord is not equal to the product of the lengths of the other chord. 4. a (6 r) r 53 5 r. ø r The radius of the tank is about. feet. b. 5 4(4 ) 5 4(6) You are 8 feet from a point of tangenc. 5. a. 6 b. The circles intersect at point (, 3), so the epicenter is at (, 3). c. center: (, 3) radius: ( ) ( 3) 5 ( ) ( 3) 5 44 t (4, 6): ( ) ( 3) 5 44 (4 ) (6 3) > 44 The point (4, 6) lies outside the range of the earthquake, so ou could not feel the earthquake. 6. a. ( ) Ï () 4()(5) () 5 6 Ï Ï 5 b. 8 p 5 p 8 5 c Ï hapter 0 Review (pp ). If a chord passes through the center of a circle, then it is called a diameter.. O n inscribed angle is an angle whose verte is on the circle and whose sides contain chords of the circle. The arc that lies in the interior of an inscribed angle and has endpoints on the angle is called the intercepted arc of the angle. opright b Mcougal Littell, a division of Houghton Mifflin ompan. 346