GEOMETRY OF THE CIRCLE TANGENTS & SECANTS

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Godfrey Griffin
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1 Geometry Of The Circle Tangents & Secants GEOMETRY OF THE CIRCLE TANGENTS & SECANTS SOLUTIONS

3 Basics Page 3 questions 1. What is the difference between a secant and a tangent? A secant passes through a circle. The two points it passes through are called intercepts. While a tangent is a straight line that touches the circle once. This point is called the point of contact.. Draw a secant and a tangent to the circle below. Label the "point of contact". F E A B D C D P Q R S DF is a tangent, E is point of contact AD is a secant, PS is also a secant 3. What s the difference between an intercept and a "point of contact"? A point of contact touches the circle once. An intercept goes through the the circle % Geometry of the Circle - Tangents and Secants K 16 1

4 Knowing More Page 6 questions 1. O is the centre of the circle below. a Show that TTUO / TTVO (SSS). b Show that OT bisects + UTV. U T O V a O is the centre In T TUO and TOVT TU = TV OU = OV OT is common ` TTUO/ TTVO ` UT = VT (Given) (Tangents from common source are equal) (Equal Radii) (SSS) (Corresponding sides of congruent triangles) b + OTU = + OTV ` OT bisects + UTV (Corresponding angles of congruent triangles). Find the length of tangent YZ if O is the centre of the circle. O 4cm 6. 4cm Y Z OY = ZY ` + ZYO = 90c ` T OYZ is right angled ` OZ = OY + YZ ` YZ = ` YZ = = cm (Tangent perpendicular to radius at point of contact) (Pythagoras) K % Geometry of the Circle - Tangents and Secants

5 Knowing More Page 7 questions 3. PR is a tangent to the circle with point of contact Q. O is the centre of the circle. + PQA = 15c and + RQB = 30c. Find reflex + AOB. Q P R A B O PR = OQ ` + OQR = + OQP = 90c ` + OQA = 90c- 15c = 75c ` + OQB = 90c- 30c = 60c ` + AQB = 75c+ 60c = 135c ` reflex + AOB = # 135c = 70c (Tangent perpendicular to radius at point of contact) (Angle at centre is twice the angle at circumference on same arc) 4. O is the centre of the circle below. AB and CD are tangents with points of contact S and T respectively. Show that + BSD = + TDS. B S A O D T C AB = ST ` + BST = 90c CD = ST ` + CTS = 90c AB ;; CD ` + BSD = + TDS (Tangent perpendicular to radius at point of contact) (Tangent perpendicular to radius at point of contact) (Alternate angles are equal) (Alternate angles on parallel lines are equal) % Geometry of the Circle - Tangents and Secants K 16 3

6 Knowing More Page 8 questions 5. MA is tangent to the larger circle and MC is tangent to the smaller circle. MB is a common tangent. O is the centre of the larger circle. Find the radius of the larger circle if MC = 19. 5cm and MO =. 1cm. A O B C M AM = OA ` + OAM = 90c AM = MB = MC ` AM = 19.5 cm ` OM = OA + AM ` OA = ` OA = = 104. cm (Tangent perpendicular to radius at point of contact) (Tangents from common source are equal) (Pythagoras) 4 K % Geometry of the Circle - Tangents and Secants

7 Using Our Knowledge Page 11 questions 1. Identify the angles equal to the labelled angles. a b c. Find the size of + LJI if IJ is tangent to the circle below. K 71c L 44c I J + LKJ = 180c-71c- 44c ` + LKJ = 65c + LKJ = + LJI ` + LJI = 65c (Sum of angles in a triangle) (Alternate segment angles) % Geometry of the Circle - Tangents and Secants K 16 5

8 Using Our Knowledge Page 1 questions 3. CE and AC are both tangents to the circle below. Find + DBC and + FDE. E D C + DBC = + BFD (Alternate segment angles) ` + DBC = 68c F 68c + FDB = + ABF ` + FDB = 87c (Alternate segment angles) + DBF = 180c-68c- 87c (Sum of angles in a triangle) 87c B ` + DBF = 5c A + FDE = + DBF ` + FDE = 5c (Alternate segment angles) 4. The circle below has centre O and tangent PQ with point of contact F. Find a + OED. b + ODF. c +EFQ. D O 30c E a + FED = + PFD ` + FED = 50c ` + OED = 50c-30c ` + OED = 0c (Alternate segment angles) P 50c F Q b + OFP = 90c ` + OFD = 90c -+ DFP (Tangent = to radius at point of contact) = 90c-50c = 40c TOFD is isosceles (OD = OF = equal radii) ` + ODF = 40c (Equal angles of isosceles T OFD ) c + EDF = + EDO + + ODF = 0c+ 40c = 60c + EDF = + EFQ (Alternate segment angles) ` + EFQ = 60c 6 K % Geometry of the Circle - Tangents and Secants

9 Using Our Knowledge Page 13 questions 5. AB and CD are tangents to the circle with points of contact P and S respectively. RS bisects + QSC. a b Show that Find the size of T PQS is an Isosceles triangle. + QRS. A P 70c 70c B D c Find the size of + RSD. a + PQS = + BPS (Alternate segment angles) Q S ` + PQS = 70c + PQS = + SPQ R ` T PQS is an Isosceles triangle C b + QSP = 180c- 140c = 40c (Sum of angles in a triangle) + DSP = + SQP (Alternate segment angles) ` + DSP = 70c + QRS = + DSQ (Alternate segment angles) + QRS = 70c+ 40c = 110c c + QSC = + SPQ (Alternate segment angles) ` + QSC = 70c RS bisects + QSC (Given) ` + RSQ = 1 ^ 70 ch ` + RSQ = 35c (RS bisects + QSC ; + QSC = 70c ) PS = QS ( T PQS ; is iscosceles) ` + PQS = + QPS = 70c (Angles opposite equal sides) ` + QSP = 180c-70c- 70c (Interior angles of T PQS ) ` + QSP = 40c + PSD = + PQS (Alternate segment angles) ` + PSD = 70c + RSD = + RSQ + + QSP+ + PSD = 35c+ 40c+ 70c = 145c % Geometry of the Circle - Tangents and Secants K 16 7

10 Thinking More Page 17 questions 1. Find x in each of the following (all measurements in cm). a A C E 0 x D AE x CD ED # # = ` 40 # x = 0 # 60 ` x = = 30 cm (Product of intercepts on intersecting chords) B b V x# TU = 3 # VT ` x# ^x+ 5h = 3# 1 (Products of intercepts of intersecting secants from external point) ` x + 5x- 36 = 0 9 U ` ^x+ 9h^x- 4h= 0 5 ` x = 4 or x =-9 3 x Since length is always positive: x = 4 cm T c Given: PQ is a tangent S x = PS# PR ` x = 5 # 9 = 5 (Square of tangent equal product of secant from common point) Q ` x = 5 16 ` x = 15 cm x 9 R P 8 K % Geometry of the Circle - Tangents and Secants

11 Thinking More Page 18 questions. Find the missing lengths in each of the following (all measurements in cm). a E D C ED = 8 BC = 7 AB = 5 Find CD = x B A CE x CB CA # # = (Products of intercepts of intersecting secants from external point) ` x ` ^x+ 8hx = 7# 1 + 8x- 84 = 0 ` ^x+ 14h^x- 6h= 0 ` x = 6 or x =-14 Since length is always positive: x = 6 cm b L M KL = 1 MN = 8 Find LM = x K N MN = x# MK (Square of tangent equal product of secant from common point) ` x ` 8 + 1x- 64 = 0 ` ^x+ 16h^x- 4h= 0 = xx ^ + 1h ` x = 4 or x =-16 Since length is always positive: x = 4 cm % Geometry of the Circle - Tangents and Secants K 16 9

12 Thinking More Page 19 questions 3. Find x and y in the diagram below: D y C x 8 4 B 6 5 A E ` CB = 4# ^x + 10h ` 8 = 4x + 40 ` 4x = 4 ` x = 6 (Square of tangent equal product of secant from common point) ` 6# x = 5# y ` 36 = 5y ` y = 36 5 = (Products of intercepts on intersecting chords) 10 K % Geometry of the Circle - Tangents and Secants

13 Thinking Even More Page 1 questions Here is a mix of more difficult problems combining all the theorems for Circle Geometry. 1. O is the centre of the circle below. PQ is a tangent with point of contact C. BCQ 30c + =. Find 5 other angles which equal 30c. A O D = = B P C 30c Q + CDB = + QCB ` + CDB = 30c ` + CAB = 30c (Alternate Segment Angle) (Angles in same segment on same arc) QP = OC (Tangent perpendicular to radius at point of contact) ` + PCO = + QCO = 90c ` + BCA = 90c- 30c = 60c ` + BDA = 60c (Right angle) (Angles in same segment on same arc) + CDA = + CBA = 90c (Angle in a semi circle) DB = OC (Line from centre to midpoint is perpendicular to chord) ` + DCO = 60c ` + DBA = 60c ` + CBD = 30c ` + CAD = 30c ` + PCD = 30c (Angles in T sum to 180c ) (Angles in same segment on same arc) (Angles in T sum to 180c ) (Angles in same segment on same arc) (Alternate segment angle) % Geometry of the Circle - Tangents and Secants K 16 11

14 Thinking Even More Page questions. BEDC is a Rhombus and GD is a tangent to the circle at E. a b Show + GEA = + BED. Show CE bisects + BED. B = = C c Show + EAB+ + EDC = 180c. A = = D G E a + GEA = + EBA + BED = + BAE (Alternate Segment Angle) (Alternate Segment Angle) AC ;; GD ` + GEA = + BAE ` + GEA = + BED (Rhombus has parallel sides) (Alternate angles are equal) b In TEBC and TEDC EB = BC = ED = DC + EBC = + EDC EC = EC ` TEBC / TEDC ` + CED = + BEC ` CE bisects + BED (Given) (Rhombus) (Common side) (SSS) (Corresponding angles of congruent triangles) c AC ;; GD EB ;; DC + BED = + EAB + BED+ + EDC = 180c ` + EAB+ + EDC = 180c (Rhombus has parallel sides) (Rhombus has parallel sides) (Alternate Segment Angle) (Supplementary cointerior angles, EB DC ) 1 K % Geometry of the Circle - Tangents and Secants

15 Thinking Even More Page 3 questions 3. In the diagram below, O is the centre of the circle and J is the point of contact of tangent KJ. Given JK = cm JP = 10. cm OP = OM = 765. cm + PNO = 37c OP = NJ and OM = NL N 37c P O M J L K a Find the length of PN. In TOPN and TOPJ OP is common ON = OJ + OPN = + OPJ = 90c ` TOPN / TOPJ ` PN = PJ = 10. cm (Equal radii) (Given) (RHS) (Corresponding sides of congruent triangles) OR OP = NJ ` OP bisects NJ (Given) (Line from centre to a chord, = to the chord, bisects it) ` NP = JP ` NP = 10. cm % Geometry of the Circle - Tangents and Secants K 16 13

16 Thinking Even More Page 3 questions J P N 37c - O M L K b Find the length of LN. In TOPN and TOMN ON is common OM = OP + OMN = + OPN = 90c ` TOPN / TOMN ` MN = PN = 10. cm (Given) (Given) (RHS) (Corresponding sides of congruent triangles) In TOML and TOMN ON = OL OM is common + OMN = + OML = 90c ` TOML / TOMN ` LM = MN = 10. ` LN = = 0.4 cm (Equal radii) (Given) (RHS) (Corresponding sides of congruent triangles) OR OM = NL ` OM bisects NL ` MN = ML ` NL = # MN = 0.4 cm (Given) (Line from centre perpendicular to chord theorem) 14 K % Geometry of the Circle - Tangents and Secants

17 Thinking Even More Page 3 questions c Find the length of LK. JK = LK # KN = LK # KN `13. 6 = LK^LK h (Square of tangent equal product of secant from common point) (Re-arranging) ` LK LK = 0 ` ^LK h^lk + 7. h= 0 (Re-arranging) ` LK = 6.8 cm (Only positive values) d Find + JOL. In TOPN / TOPJ ` + PJO = + ONP = 37c ` + JON = 180c-37c- 37c = 106c In TOPN and TOMN ON = ON OM = OP + OMN = + OPN = 90c ` TOPN / TOMN ` TOML / TOMN ` TOJN / TOLN ` + LON = + JON = 106c (Previous result) (Corresponding angles of congruent triangles) (Sum of angles in a triangle) (Common side and radii) (Given) (Given) (RHS) (RHS) (SSS) (Corresponding sides of congruent triangles) ` + JOL = 360c-106c- 106c = 148c OR TOPN / TOMN ` + PNM = + MNO = 37c ` + PNM = # 37c = 74c + JOL = # + JNL = # 74c = 148c (Angle at centre is equal to twice the angle at the circumference) % Geometry of the Circle - Tangents and Secants K 16 15

18 Thinking Even More Page 4 questions 4. JM and LM are tangents with points of contact J and L respectively. JK ML and KPM is a straight line. L K P J N M a Show + LJK = + LJM. + LJM = + JLM + JLM = + LJK ` + LJK = + LJM (Angles opposite equal sides of isosceles (Alternate + 's, JK LM ) T JML ) b Show + JML = + LKJ. Let + JLM = x ` + LJM = x ` + JML = 180c -x ( + JLM = + LJM, proved above in a ) (Sum of angles in a triangle) Also, + KJL = + MLJ = x and LKJ KJL JLM 180c = (Alternate angles, ;; LM KJ ) (Sum of angles in a triangle) ` + LKJ+ x+ x = 180c ` + LKJ = 180c -x ` + JML = + LKJ (Both equal 180c - x ) 16 K % Geometry of the Circle - Tangents and Secants

19 Thinking Even More Page 5 questions 5. In the diagram below QR and QP are tangents with points of contact R and P respectively. Let+ QPW = x; + OPV = y and + WQR = z. Q z P x y O V a Show + PTQ = + PRQ. QP = QR (Theorem 10) W T ` + PRQ = x (Isosceles triangle) but + PTQ = x ` + PTQ = + PRQ (Alternate segment angle) R U b Show TQ ;; UR. S ` + PUR = x but + PUR = + PTQ ` TQ ;; UR (Alternate segment angle) c Show + VPW = + URT = 180c. ` + VPW+ x+ y = 180c + VTP = y ` + UTR = + VTP = y + PUR = x ` + UTR+ + PUR + + URT = 180c ` y+ x+ + UTR = 180c ` y+ x+ + UTR = + VPW+ x+ y ` + UTR = + VPW (Angles on a straight line) (Vertically opposite) (Previous result) (Sum of angles in a triangle) (Substitution) d Show + URS = + WQR. + URS = + UPR (Alternate segment angle) % Geometry of the Circle - Tangents and Secants K 16 17

20 Notes 18 K % Geometry of the Circle - Tangents and Secants

21 Notes % Geometry of the Circle - Tangents and Secants K 16 19

22 Notes 0 K % Geometry of the Circle - Tangents and Secants

Answer Key. 9.1 Parts of Circles. Chapter 9 Circles. CK-12 Geometry Concepts 1. Answers. 1. diameter. 2. secant. 3. chord. 4.

Answer Key. 9.1 Parts of Circles. Chapter 9 Circles. CK-12 Geometry Concepts 1. Answers. 1. diameter. 2. secant. 3. chord. 4. 9.1 Parts of Circles 1. diameter 2. secant 3. chord 4. point of tangency 5. common external tangent 6. common internal tangent 7. the center 8. radius 9. chord 10. The diameter is the longest chord in