Given: with P in its Given:

GIVEN: In Figure 6.39a, is the perpendicular bisector of chord in PROVE: contains point O PROOF BY INDIRECT Suppose that O is not on . Draw and radii METHOD: and . [See Figure 6.39b.] Because is the perpendicular bisector of , R must be the midpoint of ; then . Also, all radii of a are . With by Identity, we have by SSS. Now by CPCTC. It follows that because these line segments meet to form congruent adjacent angles. Then is the perpendicular bisector of . But is also the perpendicular bisector of , which contradicts the uniqueness of the perpendicular bisector of a segment. Thus, the supposition must be false, and it follows that center O is on , the perpendicular bisector of chord . TV Í QR TV Í QR TV OR OR ⬜ TV ∠ORT ⬵ ∠ORV 䉭ORT ⬵ 䉭ORV OR ⬵ OR ⬵ } OT ⬵ OV TR ⬵ RV TV TV Í QR OV OT OR Í QR Í QR }O TV Í QR EXAMPLE 1 GIVEN: In Figure 6.40, has a radius of length 5 at B and FIND: CD Solution Draw radius . By the Pythagorean Theorem, According to Theorem 6.3.1, we know that ; then it follows that . CD = 2 4 = 8 CD = 2 BC BC = 4 BC 2 = 16 25 = 9 + BC 2 5 2 = 3 2 + BC 2 OC 2 = OB 2 + BC 2 OC OB = 3 OE ⬜ CD }O O C D B E Figure 6.40 Figure 6.41 쮿 CIRCLES THAT ARE TANGENT In this section, we assume that two circles are coplanar. Although concentric circles do not intersect, they do share a common center. For the concentric circles shown in Figure 6.41, the tangent of the smaller circle is a chord of the larger circle. If two circles touch at one point, they are tangent circles. In Figure 6.42, circles P and Q are internally tangent, whereas circles O and R are externally tangent. Q a P b O R Figure 6.42 Exs. 1–4 쮿 As the definition suggests, the line segment joining the centers of two circles is also commonly called the line of centers of the two circles. In Figure 6.43, or is the line of centers for circles A and B. COMMON TANGENT LINES TO CIRCLES A line segment that is tangent to each of two circles is a common tangent for these cir- cles. If the common tangent does not intersect the line of centers, it is a common exter- nal tangent. In Figure 6.44, circles P and Q have one common external tangent, ; circles A and B have two common external tangents, and . Í YZ Í WX Í ST AB Í AB a Q P T S B A b W X Z Y Figure 6.44 B A Figure 6.43 R O a D E b A D B C N M Figure 6.45 For two circles with different centers, the line of centers is the line or line segment con- taining the centers of both circles. DEFINITION Geometry in the Real World Parts and of the chain belt represent common external tangents to the circular gears. CD AB A C B D Exs. 5–6 If the common tangent does intersect the line of centers for two circles, it is a com- mon internal tangent for the two circles. In Figure 6.45, is a common internal tangent for externally tangent circles O and R; and are common internal tangents for and . }N }M Í CD Í AB Í DE GIVEN: In Figure 6.46, and are tangents to from point A PROVE: PROOF: Draw . Now and . Then because these angles have equal measures. In turn, the sides opposite and of are congruent. That is, . We apply Theorem 6.3.4 in Examples 2 and 3. AB ⬵ AC 䉭ABC ∠C ∠B ∠B ⬵ ∠C m ∠C = 1 2 mBC ¬ m ∠B = 1 2 mBC ¬ BC AB ⬵ AC }O AC AB O B C A Figure 6.46 Discover Measure the lengths of tangent segments and of Figure 6.46. How do AB and AC compare? ANSWER AC AB They are equal. The tangent segments to a circle from an external point are congruent. THEOREM 6.3.4 EXAMPLE 2 A belt used in an automobile engine wraps around two pulleys with different lengths of radii. Explain why the straight pieces named and have the same length. CD AB C A B D O O P O P C A B E D O O P O P Figure 6.47 Solution Because the pulley centered at O has the larger radius length, we extend and to meet at point E. Because E is an external point to both and , we know that and by Theorem 6.3.4. By subtracting equals from equals, . Because and , it follows that . AB = CD EC - ED = CD EA - EB = AB EA - EB = EC - ED EA = EC EB = ED }P }O CD AB 쮿 EXAMPLE 3 The circle shown in Figure 6.48 is inscribed in ; , , and . Find the lengths AM, MB, and NC. Solution Because the tangent segments from an external point are , we can let Now from from from Subtracting the second equation from the first, we have x - z = 1 y + z = 8 x + y = 9 AC = 7 x + z = 7 BC = 8 y + z = 8 AB = 9 x + y = 9 NC = CP = z BM = BN = y AM = AP = x ⬵ AC = 7 BC = 8 AB = 9 䉭ABC Discover Place three coins of the same size together so that they all touch each other. What type of triangle is formed by joining their centers? ANSWER Equilateral or Equiangular N M C A B P x x y y z z Figure 6.48 쮿 LENGTHS OF SEGMENTS IN A CIRCLE To complete this section, we consider three relationships involving the lengths of chords, secants, or tangents. The first theorem is proved, but the proofs of the remain- ing theorems are left as exercises for the student. Now we use this new equation along with the third equation on the previous page and add: Because and , . Then . Because and , , so . Summarizing, , , and . NC = 3 BM = 5 AM = 4 NC = 3 z = 3 x + z = 7 x = 4 BM = 5 y = 5 x + y = 9 x = 4 2x = 8 : x = 4 : AM = 4 x + z = 7 x - z = 1 Exs. 7–10 쮿 General Rule: With the help of auxiliary lines, Theorems 6.3.5, 6.3.6, and 6.3.7 can be proved by establishing similar triangles, followed by use of CSSTP and the Means- Extremes Property. Illustration: In the proof of Theorem 6.3.5, the auxiliary chords drawn lead to similar triangles RTV and QSV. STRATEGY FOR PROOF 왘 Proving Segment-Length Theorems in the Circle If two chords intersect within a circle, then the product of the lengths of the segments parts of one chord is equal to the product of the lengths of the segments of the other chord. THEOREM 6.3.5 Reminder AA is the method used to prove triangles similar in this section. GIVEN: Circle O with chords and intersecting at point V See Figure 6.49. PROVE: PROOF: Draw and . In and , we have vertical . Also, and are inscribed angles that intercept the same arc namely , so . By AA, . Using CSSTP, we have and so . RV VS = TV VQ RV VQ = TV VS 䉭RTV 䉭QSV ∠R ⬵ ∠Q TS ¬ ∠Q ∠R ∠s ∠1 ⬵ ∠2 䉭QSV 䉭RTV QS RT RV VS = TV VQ TQ RS V Q T R O S 1 2 Figure 6.49 Technology Exploration Use computer software if available. 1. Draw a circle with chords and intersecting at point P. See Figure 6.50. 2. Measure , , , and . 3. Show that . Answers are not “perfect.” MP PL = HP PJ PJ HP PL MP LM HJ EXAMPLE 4 In Figure 6.50, , , and . Find PM. Solution Applying Theorem 6.3.5, we have . Then PM = 2.5 8 PM = 20 4 5 = 8 PM HP PJ = LP PM LP = 8 PJ = 5 HP = 4 H J M P L Figure 6.50 쮿 쮿 In Figure 6.51, we say that secant has internal segment part and external segment part . AR RB AB EXAMPLE 5 In Figure 6.50 on page 304, , , and . Find LP and PM. Solution Because , it follows that . If and , then . Now becomes , so or or Therefore, or If , then ; conversely, if , then . That is, the segments of chord have lengths of 3 and 8. LM PM = 3 LP = 8 PM = 8 LP = 3 LP = 8 LP = 3 x = 8 x = 3 x - 8 = 0 x - 3 = 0 x - 3x - 8 = 0 x 2 - 11x + 24 = 0 24 = 11x - x 2 6 4 = x11 - x HP PJ = LP PM PM = 11 - x LP = x LM = 11 PM = LM - LP LP + PM = LM LM = 11 PJ = 4 HP = 6 Exs. 11–13 쮿 B R A T C Figure 6.51 If two secant segments are drawn to a circle from an external point, then the products of the lengths of each secant with its external segment are equal. THEOREM 6.3.6 GIVEN: Secants and for the circle in Figure 6.51 PROVE: The proof is left as Exercise 46 for the student. HINT: First use the auxiliary lines shown to prove that . 䉭 ABT 䉭ACR AB RA = AC TA AC AB EXAMPLE 6 GIVEN: In Figure 6.51, , , and FIND: AC and TA Solution Let . Because , we have , so . If and , then . The statement becomes , so or is discarded because the length of cannot be negative. Thus, , so . TA = 9 AC = 14 AC x = - 9 x = 14 or x = - 9 x + 9 = 0 x - 14 = 0 x - 14x + 9 = 0 x 2 - 5x - 126 = 0 126 = x 2 - 5x 14 9 = xx - 5 AB RA = AC TA AR = 9 BR = 5 AB = 14 TA = x - 5 AT + 5 = x AT + TC = AC AC = x TC = 5 BR = 5 AB = 14 쮿 GIVEN: Tangent and secant in Figure 6.52 PROVE: The proof is left as Exercise 47 for the student. HINT: Use the auxiliary lines shown to prove that . 䉭 TVW 䉭TXV TV 2 = TW TX TW TV If a tangent segment and a secant segment are drawn to a circle from an external point, then the square of the length of the tangent equals the product of the length of the secant with the length of its external segment. THEOREM 6.3.7 X V T W Figure 6.52 S V T R Figure 6.53 EXAMPLE 7 GIVEN: In Figure 6.53, and FIND: ST Solution If and , then . Using Theorem 6.3.7, we find that Because ST cannot be negative, . ST = 6 ST = 6 or - 6 ST 2 = 36 ST 2 = 12 3 ST 2 = SR SV SR = 12 VR = 9 SV = 3 VR = 9 SV = 3 Exs. 14–17 쮿 Exercises 6.3

1. Given: with

Find:

2. Given: and

in Find: CD

3. Given: in

and Find: RS OT = 6 OV = 9 }O OV ⬜ RS }O OE ⬜ CD OE = 6 OC = 8 mCF ¬ CD = OC OE ⬜ CD }O

4. Given:

V is the midpoint of in and Find: OR

5. Sketch two circles that have:

a No common tangents b Exactly one common tangent c Exactly two common tangents d Exactly three common tangents e Exactly four common tangents

6. Two congruent intersecting circles B and D not shown

have a line segment of centers and a common chord that are congruent. Explain why quadrilateral ABCD is a square. AC BD OT = 6 m ∠S = 15° }O RS ¬ O E C D F Exercises 1, 2 O V S R T Exercises 3, 4 In the figure for Exercises 7 to 16, O is the center of the circle. See Theorem 6.3.5.

7. Given: , ,

Find: EC

8. Given: , ,

Find: EB

9. Given: , ,

Find: DE and EC

10. Given: , ,

Find: DE and EC

11. Given: , ,

Find: CB

12. Given: , ,

Find: EC

13. Given: , ,

, and Find: x and AE

14. Given: , , ,

and Find: x and DE

15. Given: and ;

Find: DE and EC

16. Given: and ;

Find: DE and EC For Exercises 17–20, see Theorem 6.3.6.

17. Given: , ,

Find: DE AE = 15 BC = 8 AB = 6 DE :EC = 3:1 EB = 4 AE = 6 DE :EC = 2:1 EB = 8 AE = 9 EC = 6 DE = 5x 6 EB = x 3 AE = x 2 EC = 9 DE = x + 6 3 EB = 12 AE = x 2 AE = 7 BC = 4 AD = 10 AD = 8 EC = 3 AE = 6 DC = 12 EB = 5 AE = 7 DC = 16 EB = 6 AE = 8 AE = 8 EC = 5 DE = 12 DE = 8 EB = 4 AE = 6 E D A O B C Exercises 7–16 C B D E A Exercises 17–20

18. Given: , ,

Find: AD

19. Given: , ,

Find: DE

20. Given: , ,

Find: AE In the figure for Exercises 21 to 24, is tangent to the circle at S. See Theorem 6.3.7.

21. Given: and

Find: RT

22. Given: and

Find: RS

23. Given: and

Find: RS HINT: Use the Quadratic Formula. RT = 6 RS ⬵ TV TV = 6 RT = 4 RV = 12 RS = 8 RS AD = 6 BC = 6 AB = 5 AD = 3 BC = 5 AB = 4 AE = 14 AB = 6 AC = 12 S T V R Exercises 21–24

24. Given: and

Find: RT

25. For the two circles in Figures a, b, and c, find the

total number of common tangents internal and external. TV = 9 RT = 1 2 RS a b c

26. For the two circles in Figures a, b, and c, find the

total number of common tangents internal and external. a b c In Exercises 27 to 30, provide a paragraph proof.

27. Given: and

are tangent at point F Secant to Secant to Common internal tangent Prove: AC AB = AE AD AF }Q AE }O AC }Q }O O Q A C D F B E

28. Given: with and

Prove: is isosceles 䉭ABC OM ⬵ ON ON ⬜ BC OM ⬜ AB }O N B A C O M

29. Given: Quadrilateral

ABCD is circumscribed about Prove:

30. Given: in

Prove:

31. Does it follow from Exercise 30

that is also congruent to ? What can you conclude about and in the drawing? What can you conclude about and ?

32. In not shown,

is a diameter and T is the midpoint of semicircle . What is the value of the ratio ? The ratio ?

33. The cylindrical brush

on a vacuum cleaner is powered by an electric motor. In the figure, the drive shaft is at point D. If find the measure of the angle formed by the drive belt at point D; that is, find .

34. The drive mechanism on a treadmill is powered by an

electric motor. In the figure, find if is 36° larger than

35. Given: Tangents , , and to at

points M, N, and P, respectively , , Find: AM, PC, and BN AC = 12 BC = 16 AB = 14 }O AC BC AB mAC ¬. m ABC ២ m ∠D m ∠D mAC ¬ = 160°, RT RO RT RS RTS ២ RS }O EB DE CE AE 䉭CBE 䉭ADE 䉭ABD ⬵ 䉭CDB }P AB ⬵ DC DA + BC AB + CD = }O C A D B E P Exercises 30, 31 C A D B C A D B Exercises 33, 34 C D B A O N M C A B P O B A O D P Exercises 38, 39 Exercises 42, 43

36. Given: is inscribed in

isosceles right The perimeter of is Find: TM 8 + 4 22 䉭RST 䉭RST }Q N M R S P Q T

37. Given: is an external

tangent to and at points A and B ; radii lengths for and are 4 and 9, respectively Find: AB HINT: The line of centers contains point C, the point at which and are tangent.

38. The center of a circle of radius 3 inches is at a distance

of 20 inches from the center of a circle of radius 9 inches. What is the exact length of common internal tangent HINT: Use similar triangles to find OD and DP. Then apply the Pythagorean Theorem twice. AB ? }Q }O OQ }Q }O }Q }O AB B A O Q C

39. The center of a circle of radius 2 inches is at a distance

of 10 inches from the center of a circle of radius length 3 inches. To the nearest tenth of an inch, what is the approximate length of a common internal tangent? Use the hint provided in Exercise 38.

40. Circles O, P, and Q are

tangent as shown at points X, Y, and Z. Being as specific as possible, explain what type of triangle is if: a , , b , ,

41. Circles O, P, and Q are tangent as shown at points X, Y,

and Z. Being as specific as possible, explain what type of triangle is if: a , , b , ,

42. If the larger gear has 30 teeth and the smaller gear has

18, then the gear ratio larger to smaller is 5:3. When the larger gear rotates through an angle of 60°, through what angle measure does the smaller gear rotate? QZ = 2 PY = 2 OX = 2 QZ = 1 PY = 4 OX = 3 䉭PQO QZ = 2 PY = 3 OX = 2 QZ = 1 PY = 3 OX = 2 䉭PQO O Q P X Y Z Exercises 40, 41