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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 of Circles (2)

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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.

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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

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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.

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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:

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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

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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

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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

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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)

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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.

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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

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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

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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

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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

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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.

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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)

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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

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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

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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)

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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)

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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)

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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)

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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)

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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

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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.

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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

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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

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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

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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

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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

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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

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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

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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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