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

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Presentation on theme: "8 8.1Tangents to a Circle 8.2Tangents to a Circle from an External Point Chapter Summary Case Study 8.3Angles in the Alternate Segments Basic Properties."— Presentation transcript:

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

2 P. 2 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. Can you give me one real-life example of a circle and a straight line? Yes, the wheel of a train is a circle and the rail is a straight line.

3 P Tangents to a Circle 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. 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. 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 P Tangents to a Circle There is a close relationship between the tangent to a circle and the radius joining the point of contact: 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) This theorem can be proved by contradiction: Suppose AB is not perpendicular to the radius OT. Then we can find another point T on AB such that OT  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 P Tangents to a Circle The converse of Theorem 8.1 is also true: 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. Hence we can deduce an important fact:

6 P Tangents to a Circle Example 8.1T Solution: 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. (a)OT  OC  9 cm (radii) ∴  OCT is an equilateral triangle. ∴  COT   OTC  60  (prop. of equilateral  )  OTA  90  (tangent  radius) In  OAT,  CAT   OTA   COT  180  (  sum of  ) ∴  CTA  90   60   30   CAT  30 

7 P Tangents to a Circle Example 8.1T Solution: 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. (b) ∵  CTA   CAT  30  (proved in (a)) ∴ CA  CT  9 cm (sides opp. equal  s) In  OAT, AT 2  OT 2  OA 2 (Pyth. Theorem) AT  cm  cm

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

9 P 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  OTA   OTB (RHS) :  OAT   OBT  90  (tangent  radius) OT  OT (common side) OA  OB (radii) Hence the corresponding sides and the corresponding angles of  OTA and  OTB are equal: TA  TB (corr. sides,   s)  TOA   TOB (corr.  s,   s)  OTA   OTB (corr.  s,   s)

10 P Tangents to a Circle from an External Point Properties of tangents 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.

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

12 P Tangents to a Circle from an External Point Example 8.4T Solution: 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. ((  ABC  66  180  (opp.  s, cyclic quad.)  ABC  114   ACB :  BAC  AB : BC (arcs prop. to  s at ⊙ ce ) (( x :  BAC  1 : 2 ∴  BAC  2x In  ABC,  ABC   BAC  x  180  (  sum of  ) 114   2x  x  180  x  22  2x2x

13 P Tangents to a Circle from an External Point Example 8.4T Solution: 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. ((  AOC  2  66  (  at the centre twice  at ⊙ ce )  OAT   OCT  90  (tangent  radius)  132  ∴  AOC   OAT   ATC   OCT  360  (  sum of polygon) 132   90    ATC  90   360   ATC  48  ∵ TC  TA (tangent properties) ∴  TCA   TAC (base  s, isos.  ) In  TAC,  ATC  2  TAC  180  (  sum of  )  TAC  66  ∵  BAC  2x  44  ∴ y  22  2x2x

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

15 P 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. Then a is an angle in the alternate segment with respect to b. Notes: We can construct infinity many angles in the alternate segment with respect to b.

16 P Angles in the Alternate Segments The figure shows another angle in the alternate segment with respect to b with BR passing through the centre O.  R   C  a (  in the same segment)  RAB  90  (  in semicircle) In  ABR,  R   RAB   ABR  180  (  sum of  )  ABR  90   a ∵  ABR   ABQ  90  (tangent  radius) ∴ (90   a)  b  90  a  b a  90    ABR  180  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)

17 P Angles in the Alternate Segments Example 8.6T Solution: 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. (a) ∵ BA  BT (given) ∴  BAT   BTA (base  s, isos.  )  48  ∴  ACB   BAT (  in alt. segment)  48  (b)  CBA   BTA   BAT (ext.  of  )  96  ∴  CAS   CBA (  in alt. segment)  96 

18 P Angles in the Alternate Segments Example 8.7T Solution: The figure shows an inscribed circle of  ABC. 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  PQR. ∵ AP  AR (tangent properties) ∴  APR   ARP (base  s, isos.  ) In  PAR, 40    APR   ARP  180  (  sum of  )  ARP  70  Similarly, ∵ CQ  CR (tangent properties) ∴  CRQ   CQR  56  ∴  PQR   ARP  70  (  in alt. segment) ∴  QPR   CRQ  56  (  in alt. segment) ∴  PRQ  180   70   56   54 

19 P Angles in the Alternate Segments The converse of Theorem 8.4 is also true: 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)

20 P Angles in the Alternate Segments Example 8.8T Solution: In the figure, AB // PQ and CD is a common chord of the circles. Prove that PQ is a tangent to the larger circle.  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 P Tangents to a Circle Chapter Summary 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 P. 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 P. 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)

24 Follow-up Tangents to a Circle In the figure, O is the centre of the circle. PT is a diameter to the circle. BT  8 cm, PT  12 cm and OB  10 cm. (a)Find  OTB. Hence show that AB is a tangent to the circle. (b)Find the length of BP. (Express the answer in surd form.) (a)OT  6 cm Solution: In  OBT, BT 2  OT 2  (8 2  6 2 ) cm 2  100 cm 2  OB 2 ∴  OTB  90  (converse of Pyth. Theorem) ∴ AB is a tangent to the circle at T. (converse of tangent  raduis) (b)BP  cm (Pyth. Theorem)  cm

25 Follow-up 8.2 Solution: 8.1 Tangents to a Circle In the figure,  ABC is an equilateral triangle and BD is a tangent to the circle. Find  CBD. (Hint: Join B and C to the centre of the circle respectively.)  BOC  2   BAC (  at the centre twice  at ⊙ ce ) Join OB and OC. In  OBC, ∵ OB  OC (radii) ∴  OBC   OCB (base  s, isos.  ) O  2  60  (prop. of equilateral  )  120  ∴ 120   2  OBC  180  (  sum of  )  OBC  30  ∵  OBD  90  (tangent  radius) ∴  CBD  60  By ‘equal chords, equal  s’, we can also find  BOC.

26 Follow-up 8.3 Solution: In the figure, TA and TB are tangents to the circle with centre O. C is a point on the circle such that  ACB  68 , find x and y.  AOB  2   ACB (  at the centre twice  at ⊙ ce ) ∴  AOT   BOT  68  (tangent properties) In  OAT,  AOT   OAT  x  180  (  sum of  ) 8.2 Tangents to a Circle from an External Point  2  68   OAT  90  (tangent  radius) 68   90   x  180  x  22 

27 Follow-up 8.3 Solution: In the figure, TA and TB are tangents to the circle with centre O. C is a point on the circle such that  ACB  68 , find x and y.  OTB  x (tangent properties) 8.2 Tangents to a Circle from an External Point ∵ TA  TB (tangent properties) ∴  TAB  y (base  s, isos.  ) In  ATB,  ATB   TAB  y  180  (  sum of  ) 2x  y  y  180  2(22  )  2y  180  y  68 

28 Follow-up 8.4 Solution: In the figure, a circle is inscribed in  ABC. The circle touches AB, BC and CA at P, Q and R respectively. If  BAC = 70  and  QRC = 50 , find  PRQ and  BPQ. 8.2 Tangents to a Circle from an External Point In  PAR, AP  AR (tangent properties) ∴ 70    ARP   APR  180  (  sum of  ) ∴  APR   ARP (base  s, isos.  )  ARP  55  ∴  ARP   PRQ  50  180  (adj.  s on st. line)  PRQ  75 

29 Follow-up 8.4 Solution: In the figure, a circle is inscribed in  ABC. The circle touches AB, BC and CA at P, Q and R respectively. If  BAC  70  and  QRC  50 , find  PRQ and  BPQ. 8.2 Tangents to a Circle from an External Point In  QCR,CQ  CR (tangent properties) ∴ 2(50  )   QCR  180  (  sum of  ) ∴  CQR   CRQ  50  (base  s, isos.  )  QCR  80  In  ABC,  A   B   C  180  (  sum of  ) ∴  B  30  In  BPQ,BP  BQ (tangent properties) ∴  BPQ   BQP (base  s, isos.  ) ∴ 2  BPQ  30   180  (  sum of  )  BPQ  75 

30 Follow-up Tangents to a Circle from an External Point Solution: In the figure, a circle is inscribed in  ABC and it touches AB, BC and CA at P, Q and R respectively. If AB  12 cm, BC  18 cm and CA  14 cm, find the length of BQ. AP  AR, BP  BQ and CQ  CR. (tangent properties) Let AP  AR  a, BP  BQ  b and CQ  CR  c. Then a  b  12 cm, b  c  18 cm and c  a  14 cm. ∴ BQ  8 cm Since the perimeter of  ABC  44 cm, i.e., (a  b)  (b  c)  (c  a)  44 cm 2a  2b  2c  44 cm a  b  c  22 cm

31 Follow-up Angles in the Alternate Segments Solution: In the figure, PQ is a tangent to the circle. If  ABC  105  and  DAC  50 , find  DAP.  CAP   CBA (  in alt. segment) ∴  DAP  105   50   55   105 

32 Follow-up Angles in the Alternate Segments Solution: In the figure, TA and TB are tangents to the circle with centre O.  ATB  42  and  CBS  65 . Find  ACB and  BOC. ∵ TA  TB (tangent properties) ∴  TAB   TBA (base  s, isos.  ) In  ATB, 42    TAB   TBA  180  (  sum of  )  TAB  69  ∴  ACB   TAB  69  (  in alt. segment)  BAC   SBC (  in alt. segment)  65   BOC  2   BAC (  at the centre twice  at ⊙ ce )  130 

33 Follow-up Angles in the Alternate Segments Solution: In the figure, TA is a tangent to the circle. AT  CT,  ABC  68  and  ATC  44 . (a)Find  ACT. (b)Prove that TC is also a tangent to the circle. (a)  CAT   ABC (  in alt. segment) ∴ TC is a tangent to the larger circle. (converse of  in alt. segment)  68  In  ATC, 44    CAT   ACT  180  (  sum of  ) (b) ∵  TCA   ABC  ACT  68 


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