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Basic Properties of Circles (2)

8. Basic Properties of Circles (2). Case Study. 8.1 Tangents to a Circle. 8.2 Tangents to a Circle from an External Point. 8.3 Angles in the Alternate Segments. Chapter Summary. Yes, the wheel of a train is a circle and the rail is a straight line. Can you give me one

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Basic Properties of Circles (2)

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  1. 8 Basic Properties of Circles (2) Case Study 8.1 Tangents to a Circle 8.2 Tangents to a Circle from an External Point 8.3 Angles in the Alternate Segments Chapter Summary

  2. Yes, the wheel of a train is a circle and the rail is a straight line. Can you give me one real-life example of a circle and a straight line? Case Study The wheels of a train and the rails illustrate an important geometrical relationship between circles and straight lines. When the train travels on the rails, it shows how a circle and a straight line touch each other at only one point.

  3. 8.1 Tangents to a Circle In case 3, at each point on a circle, we can draw exactly one straight line such that the line touches the circle at exactly one point. We can draw a straight line AB and a circle in three different ways: Case 1: The straight line does not meet the circle. Case 2: The straight line cuts the circle at two distinct points, P and Q. Case 3: The straight line touches the circle at exactly one point, T. Tangent to a circle: straight line if and only if touching the circle at exactly one point Point of contact (point of tangency): point common to both the circle and the straight line

  4. 8.1 Tangents to a Circle Theorem 8.1 If AB is a tangent to the circle with centre O at T, then AB is perpendicular to the radius OT. Symbolically, ABOT. (Reference: tangent  radius) There is a close relationship between the tangent to a circle and the radius joining the point of contact: This theorem can be proved by contradiction: Suppose AB is not perpendicular to the radius OT. ThenwecanfindanotherpointT onABsuchthatOTAB. Using Pythagoras’ Theorem, OT is shorter than OT. Thus Tlies inside the circle. ∴ AB cuts the circle at more than one point.

  5. 8.1 Tangents to a Circle Theorem 8.2 OT is a radius of the circle with centre O and AB is a straight line that intersects the circle at T. If AB is perpendicular to OT, then AB is a tangent to the circle at T. In other words, if ABOT, then AB is a tangent to the circle at T. (Reference: converse of tangent  radius) The perpendicular to a tangent at its point of contact passes through the centre of the circle. The converse of Theorem 8.1 is also true: Hence we can deduce an important fact:

  6. 8.1 Tangents to a Circle ∴ CTA  90  60  30 CAT  30 Example 8.1T In the figure, O is the centre of the circle. AB is a tangent to the circle at T. OCTC 9 cm.   (a) Find CAT and CTA. (b) Find the length of AT. Solution: (a) OTOC 9 cm(radii) ∴ DOCT is an equilateral triangle. ∴ COT  OTC  60 (prop. of equilateral D) OTA  90 (tangent  radius) In DOAT, CATOTA COT  180 ( sum of D)

  7. 8.1 Tangents to a Circle AT cm cm Example 8.1T In the figure, O is the centre of the circle. AB is a tangent to the circle at T. OCTC 9 cm.   (a) Find CAT and CTA. (b) Find the length of AT. Solution: (b) ∵ CTA CAT  30(proved in (a)) ∴ CACT 9 cm (sides opp. equal s) In DOAT, AT2OT2 OA2(Pyth. Theorem)

  8. 8.1 Tangents to a Circle TBQ40 Example 8.2T In the figure, AB is a tangent to the circle at T. POQB is a straight line. If ATP 65, find TBQ. Solution: Join OT. OTA  90 (tangent  radius) ∴OTP  90  65  25 ∵ OPOT (radii) ∴ OPT OTP  25 (base s, isos. D) In DBPT, ATP TBQOPT (ext.  of D) 65 TBQ25

  9. 8.2 Tangents to a Circle from an External Point Consider an external point T of a circle. We can always draw two tangents from that point. In the figure, we can prove that DOTADOTB (RHS): OAT  OBT  90 (tangent  radius) OT  OT(common side) OA  OB(radii) Hence the corresponding sides and the corresponding angles of DOTA and DOTB are equal: TA  TB(corr. sides, Ds) TOA  TOB(corr. s, Ds) OTA  OTB(corr. s, Ds)

  10. 8.2 Tangents to a Circle from an External Point Theorem 8.3 In the figure, if TA and TB are the two tangents drawn to the circle with centre O from an external point T, then (a) the lengths of the two tangents are equal, that is, TATB; (b) the two tangents subtend equal angles at the centre, that is, TOATOB; (c) the line joining the external point to the centre of the circle is the angle bisector of the angle included by the two tangents, that is, OTAOTB. (Reference: tangent properties) In the figure, OT is the axis of symmetry. Properties of tangents from an external point:

  11. 8.2 Tangents to a Circle from an External Point ATB  50 AOB  130 Example 8.3T In the figure, TA and TB are tangents to the circle with centre O. If ABT 65, find (a) ATB, (b) AOB. Solution: (a) ∵ TA  TB(tangent properties) ∴ TAB TBA (base s, isos. D)  65 In DTAB, ATB 2(65)  180 ( sum of D) (b) OAT  OBT  90 (tangent  radius) ∴ AOB OAT ATB OBT  360( sum of polygon) AOB 90 50 90  360

  12. 8.2 Tangents to a Circle from an External Point In the figure, TA and TC are tangents to the circle with centre O. If AB : BC 1 : 2 and ADC 66, find x and y. ( ( 2x ( ( ACB : BAC  AB : BC (arcs prop. to s at ⊙ce) x 22 Example 8.4T Solution: ABC 66  180 (opp. s, cyclic quad.) ABC  114 x : BAC 1 : 2 ∴ BAC2x In DABC, ABC BAC x 180 ( sum of D) 114 2x x 180

  13. 8.2 Tangents to a Circle from an External Point In the figure, TA and TC are tangents to the circle with centre O. If AB : BC 1 : 2 and ADC 66, find x and y. ( ( 2x ∴y22 Example 8.4T Solution: AOC  2  66( at the centre twice at ⊙ce)  132 OAT  OCT  90(tangent  radius) ∴ AOC OAT ATC OCT  360( sum of polygon) 13290ATC 90  360 ATC  48 ∵ TC  TA(tangent properties) ∴ TCA  TAC (base s, isos. D) In DTAC, ATC 2TAC  180 ( sum of D) ∵ BAC 2x  44 TAC  66

  14. 8.2 Tangents to a Circle from an External Point  56 cm Example 8.5T The figure shows an inscribed circle in a quadrilateral ABCD. If AB 16 cm and CD 12 cm, find the perimeter of the quadrilateral. S Solution: P R Referring to the figure, AP AS, BP BQ, CQ CR and DR DS. (tangent properties) Q Let AP  AS  a, BP  BQ  b, CQ  CR  c and DR  DS  d. Then ab 16 cm and cd 12 cm. ∵ DA AS  SD and BC BQ  QC  a  d  b  c ∴ Perimeter  16 cm (b  c) 12 cm (a  d)  16 cm 12 cm a  b  c  d

  15. 8.3 Angles in the Alternate Segments In the figure, AB is a tangent to the circle at T and PT is a chord of the circle. Tangent-chord angles: angles formed between a chord and a tangent to a circle, such as PTA andPTB. Alternate segment: segment lying on the opposite side of a tangent-chord angle  segment I is the alternate segment with respect to PTB  segment II is the alternate segment with respect to PTA Consider the tangent-chord angle b. Thenaisanangleinthealternatesegmentwithrespecttob. Notes: We can construct infinity many angles in the alternate segment with respect to b.

  16. 8.3 Angles in the Alternate Segments Theorem 8.4 A tangent-chord angle of a circle is equal to any angle in the alternate segment. Symbolically, a  b and p  q. (Reference:  in alt. segment) The figure shows another angleinthealternatesegmentwithrespecttob with BR passing through the centre O. R  C  a( in the same segment) RAB  90 ( in semicircle) In DABR, R  RAB ABR  180 ( sum of D) a  90 ABR  180 ABR  90  a ∵ ABR  ABQ  90 (tangent  radius) ∴ (90a)b 90 a b

  17. 8.3 Angles in the Alternate Segments ∴ ACB  BAT ( in alt. segment)  48 ∴ CAS CBA ( in alt. segment)  96 Example 8.6T In the figure, TS is a tangent to the circle. TBC is a straight line. BABT and ATB 48. (a) Find ACB. (b) Find CAS. Solution: (a) ∵ BA  BT (given) ∴ BAT  BTA (base s, isos. D)  48 (b) CBA  BTA BAT (ext.  of D)  96

  18. 8.3 Angles in the Alternate Segments ∴ PQR ARP 70( in alt. segment) ∴ PRQ  180 70  56  54 ∴ QPR CRQ 56 ( in alt. segment) Example 8.7T The figure shows an inscribed circle of DABC. The circle touches the sides of the triangle at P, Q and R respectively. If BAC 40 and ACB 68, find all the angles in DPQR. Solution: ∵ AP  AR(tangent properties) ∴ APR ARP (base s, isos. D) In DPAR, 40  APR ARP  180 ( sum of D) ARP  70 Similarly, ∵ CQ  CR(tangent properties) ∴ CRQ CQR 56

  19. 8.3 Angles in the Alternate Segments Theorem 8.5 A straight line is drawn through an end point of a chord of a circle. If the angle between the straight line and the chord is equal to an angle in the alternate segment, then the straight line is a tangent to the circle. In other words, if xy, then TA is a tangent to the circle at A. (Reference: converse of  in alt. segment) The converse of Theorem 8.4 is also true:

  20. 8.3 Angles in the Alternate Segments Example 8.8T In the figure, AB // PQ and CD is a common chord of the circles. Prove that PQ is a tangent to the larger circle. Solution: BAC  CQP (alt. s, AB // PQ) BAC  CDQ (ext. , cyclic quad.) ∴ CQP  CDQ ∴ PQ is a tangent to the larger circle. (converse of  in alt. segment)

  21. Chapter Summary 8.1 Tangents to a Circle 1. If AB is a tangent to the circle with centre O at T, then AB is perpendicular to the radius OT. (Ref: tangent  radius) 2. OT is a radius of the circle with centre O and ATB is a straight line. If AB is perpendicular to OT, then AB is a tangent to the circle at T. (Ref: converse of tangent  radius)

  22. Chapter Summary 8.2 Tangents to a Circle from an External Point If TA and TB are tangents to the circle with centre O, from an external point T, then (a) TATB; (The length of the two tangents are equal.) (b) TOATOB; (Two tangents subtend equal angles at the centre.) (c) OTAOTB. (OT bisects the angle between the two tangents.) (Ref: tangent properties)

  23. Chapter Summary 8.3 Angles in the Alternate Segments 1. If TA is a tangent to the circle, then x  y and p  q. (Ref: in alt. segment) 2. If x  y, then TA is a tangent. (Ref: converse of  in alt. segment)

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