Common Tangents and Tangent Circles

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1 Common Tangents and Tangent Circles CK12 Editor Say Thanks to the Authors Click (No sign in required)

2 To access a customizable version of this book, as well as other interactive content, visit AUTHOR CK12 Editor CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform. Copyright 2012 CK-12 Foundation, The names CK-12 and CK12 and associated logos and the terms FlexBook and FlexBook Platform (collectively CK-12 Marks ) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution/Non- Commercial/Share Alike 3.0 Unported (CC BY-NC-SA) License ( as amended and updated by Creative Commons from time to time (the CC License ), which is incorporated herein by this reference. Complete terms can be found at Printed: March 1, 2013

3 Concept 1. Common Tangents and Tangent Circles CONCEPT 1 Common Tangents and Tangent Circles Learning Objectives Solve problems involving common internal tangents of circles. Solve problems involving common external tangents of circles. Solve problems involving externally tangent circles. Solve problems involving internally tangent circles. Common tangents to two circles may be internal or external. A common internal tangent intersects the line segment connecting the centers of the two circles whereas a common external tangent does not. Common External Tangents Here is an example in which you might encounter the use of common external tangents. Example 1 Find the distance between the centers of the circles in the figure. 1

4 Let s label the diagram and draw a line segment that joins the centers of the two circles. Also draw the segment AE perpendicular the radius BC. Since DC is tangent to both circles, DC is perpendicular to both radii: AD and BC. We can see that AECD is a rectangle, therefore EC = AD = 15 in. This means that BE = 25 in 15 in = 10 in. ABE is a right triangle with AE = 40 in and BE = 10 in. We can apply the Pythagorean Theorem to find the missing side, AB. (AB) 2 = (AE) 2 + (BE) 2 (AB) 2 = AB = in The distance between the centers is approximately 41.2 inches. Common Internal Tangents Here is an example in which you might encounter the use of common internal tangents. Example 2 AB is tangent to both circles. Find the value of x and the distance between the centers of the circles. 2

5 Concept 1. Common Tangents and Tangent Circles AC AB Tangent is perpendicular to the radius BD AB Tangent is perpendicular to the radius CAE = DBE Both equal 90 CEA = BED Vertical angles CEA BED AA similarity postulate Therefore, Using the Pythagorean Theorem on CEA : AC BD = AE EB 5 x = 8 12 x = 5 12 x = (CE) 2 = (AC) 2 + (AE) 2 (CE) 2 = = 89 CE 9.43 Using the Pythagorean Theorem on BED : (DE) 2 = (BD) 2 + (DE) 2 (DE) 2 = = BE The distance between the centers of the circles is CE + DE Two circles are tangent to each other if they have only one common point. Two circles that have two common points are said to intersect each other. Two circles can be externally tangent if the circles are situated outside one another and internally tangent if one of them is situated inside the other. 3

6 Externally Tangent Circles Here are some examples involving externally tangent circles. Example 3 Circles tangent at T are centered at M and N. Line ST is tangent to both circles at T. Find the radius of the smaller circle if SN SM. ST T M tangent is perpendicular to the radius. ST T N tangent is perpendicular to the radius. In the right triangle ST N,cos35 = 9 SN SN = 9 cos We are also given that SN SM. Therefore, m MSN = 90 m SMT = = 55. Also, m ST N = 90 m T SN = = 55. Therefore, SNM SNT by the AA similarity postulate. SN MN = T N SN 11 T M + 9 = (T M + 9) = 121 9T M + 81 = 121 9T M = 40 T M 4.44 The radius of the smaller circle is approximately Example 4 Two circles that are externally tangent have radii of 12 inches and 8 inches respectively. Find the length of tangent AB. 4

7 Concept 1. Common Tangents and Tangent Circles Label the figure as shown. In DOQ,OD = 4 and OQ = 20. Therefore, (DQ) 2 = (OQ) 2 (OD) 2 (DQ) 2 = = 384 DQ 19.6 CB AC QB tangent is perpendicular to the radius. AC OC tangent is perpendicular to the radius. Therefore, OCA = QBA both equal 90 OAC = QAB same angle. AOC AQB by the AA similarity postulate. Therefore, QB OC = AB AC 8 12 = AB 8(AB ) = 12AB AB AB AB 4AB AB Internally Tangent Circles Here is an example involving internally tangent circles. Example 5 5

8 Two diameters of a circle of radius 15 inches are drawn to make a central angle of 48. A smaller circle is placed inside the bigger circle so that it is tangent to the bigger circle and to both diameters. What is the radius of the smaller circle? OA and OB are two tangents to the smaller circle from a common point so by Theorem 9-3, ON bisects m NOB = 24. In ONB we use sin24 = NB ON ON = sin NB 2.46 NB. 24 Draw CD from the points of tangency between the circles perpendicular to OD. AOB In OCD we use sin24 = CD OC CD = sin 24 (OC) 0.41(15) 6.1. We also have OB NB because a tangent is perpendicular to the radius. Therefore, OBN = ODC both equal 90 6

9 Concept 1. Common Tangents and Tangent Circles COD = BON same angle. Therefore, ONB OCD by the AA similarity postulate. This gives us the ratio ON OC = NB CD. ON = OC NC = 15 NB (NB = NC, since they are both radii of the small circle). 15 NB = NB 6.1(15 NB) = 15 NB NB = 15 NB = 21.1 NB NB = 4.34 Lesson Summary In this section we learned about externally and internally tangent circles. We looked at the different cases when two circles are both tangent to the same line, and/or tangent to each other. Review Questions CD is tangent to both circles. 1. AC = 8,BD = 5,CD = 12. Find AB. 2. AB = 20,AC = 15 and BD = 10. Find CD. 3. AB = 24,AC = 18 and CD = 19. Find BD. 4. AB = 12,CD = 16 and BD = 6. Find AC. AC is tangent to both circles. Find the measure of angle CQB. 7

10 5. AO = 9 and AB = BQ = 20 and BC = BO = 18,AO = 9 8. CB = 7,CQ = 5 For 9 and 10, find x. 9. DC = 2x + 3; EC = x DC = 4x 9; EC = 2x + 21 Circles tangent at T are centered at M and N. ST is tangent to both circles at T. Find the radius of the smaller circle if SN SM SM = 22,T N = 25, SNT = 40

11 Concept 1. Common Tangents and Tangent Circles 12. SM = 23,SN = 18, SMT = Four identical coins are lined up in a row as shown. The distance between the centers of the first and the fourth coin is 42 inches. What is the radius of one of the coins? 14. Four circles are arranged inside an equilateral triangle as shown. If the triangle has sides equal to 16 cm, what is the radius of the bigger circle? What are the radii of the smaller circles? 15. In the following drawing, each segment is tangent to each circle. The largest circle has a radius of 10 inches. The medium circle has a radius of 8 inches. What is the radius of the smallest circle tangent to the medium circle? 16. Circles centered at A and B are tangent at T. Show that A, B and T are collinear. 17. TU is a common external tangent to the two circles. VW is tangent to both circles. Prove that TV =VU =VW. 18. A circle with a 5 inch radius is centered at A, and a circle with a 12 inch radius is centered at B, where A and B are 17 inches apart. The common external tangent touches the small circle at P and the large circle at Q. What kind of quadrilateral is PABQ? What are the lengths of its sides? 9

12 Review Answers ; Proof 17. Proof 18. Right trapezoid; AP = 5;BQ = 12;AB = 17;PQ =