Basic Properties of Circles (2)

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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

Case Study 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? 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.

8.1 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

8.1 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.

8.1 Tangents to a Circle The converse of Theorem 8.1 is also true:
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) Hence we can deduce an important fact: The perpendicular to a tangent at its point of contact passes through the centre of the circle.

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

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

Example 8.2T 8.1 Tangents to a Circle Solution:
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 TBQ  40

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)

8.2 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.

8.2 Tangents to a Circle from an External Point
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) ATB  50 (b) OAT  OBT  90 (tangent  radius) ∴ AOB  OAT  ATB  OBT  360 ( sum of polygon) AOB  90  50  90  360 AOB  130

8.2 Tangents to a Circle from an External Point
Example 8.4T 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 Solution: 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 DABC, ABC  BAC  x  180 ( sum of D) 114  2x  x  180 x  22

8.2 Tangents to a Circle from an External Point
Example 8.4T 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 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 ∴ y  22

8.2 Tangents to a Circle from an External Point
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  56 cm

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. 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.

8.3 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 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 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)

8.3 Angles in the Alternate Segments
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 ∴ ACB  BAT ( in alt. segment)  48 (b) CBA  BTA  BAT (ext.  of D)  96 ∴ CAS  CBA ( in alt. segment)  96

8.3 Angles in the Alternate Segments
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 ∴ PQR  ARP  70 ( in alt. segment) Similarly, ∵ CQ  CR (tangent properties) ∴ CRQ  CQR  56 ∴ PRQ  180  70  56  54 ∴ QPR  CRQ  56 ( in alt. segment)

8.3 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)

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)

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)

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)

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)

Follow-up 8.1 8.1 Tangents to a Circle Solution:
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.) Solution: (a) OT  6 cm In DOBT, BT 2  OT 2  (8 2  6 2 ) cm 2  100 cm2  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

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

Follow-up 8.3 8.2 Tangents to a Circle from an External Point
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. Solution: AOB  2  ACB ( at the centre twice  at ⊙ce)  2  68 ∴ AOT  BOT  68 (tangent properties) OAT  90 (tangent  radius) In DOAT, AOT  OAT  x  180 ( sum of D) 68  90  x  180 x  22

Follow-up 8.3 8.2 Tangents to a Circle from an External Point
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. Solution: OTB  x (tangent properties) ∵ TA  TB (tangent properties) ∴ TAB  y (base s, isos. D) In DATB, ATB  TAB  y  180 ( sum of D) 2x  y  y  180 2(22)  2y  180 y  68

Follow-up 8.4 8.2 Tangents to a Circle from an External Point
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. Solution: In DPAR, AP  AR (tangent properties) ∴ APR  ARP (base s, isos. D) ∴ 70  ARP  APR  180 ( sum of D) ARP  55 ∴ ARP  PRQ  50  180 (adj. s on st. line) PRQ  75

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

Follow-up 8.5 8.2 Tangents to a Circle from an External Point
In the figure, a circle is inscribed in DABC 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. Solution: 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. Since the perimeter of DABC  44 cm, i.e., (a  b)  (b  c)  (c  a)  44 cm 2a  2b  2c  44 cm a  b  c  22 cm ∴ BQ  8 cm

Follow-up 8.6 8.3 Angles in the Alternate Segments Solution:
In the figure, PQ is a tangent to the circle. If ABC  105 and DAC  50, find DAP. Solution: CAP  CBA ( in alt. segment)  105 ∴ DAP  105  50  55

Follow-up 8.7 8.3 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. Solution: ∵ TA  TB (tangent properties) ∴ TAB  TBA (base s, isos. D) In DATB, 42  TAB  TBA  180 ( sum of D) TAB  69 ∴ ACB  TAB  69 ( in alt. segment) BAC  SBC ( in alt. segment)  65 BOC  2  BAC ( at the centre twice  at ⊙ce)  130

Follow-up 8.8 8.3 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. Solution: (a) CAT  ABC ( in alt. segment)  68 In DATC, 44  CAT  ACT  180 ( sum of D) ACT  68 (b) ∵ TCA  ABC ∴ TC is a tangent to the larger circle. (converse of  in alt. segment)