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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

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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.

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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

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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.

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**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.

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**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

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**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

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**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

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**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)

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**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.

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**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

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**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

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**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 2TAC 180 ( sum of D) ∵ BAC 2x 44 TAC 66 ∴ y 22

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**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

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**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.

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**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)

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**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

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**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)

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**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)

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**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)

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**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)

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**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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**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 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

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**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 2OBC 180 ( sum of D) OBC 30 ∵ OBD 90 (tangent radius) ∴ CBD 60

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**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

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**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

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**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

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**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) ∴ 2BPQ 30 180 ( sum of D) BPQ 75

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**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

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**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

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**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

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**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)

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