1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection. We have two circles with centres O and O' respectively such that they intersect each other at P and Q. To prove that ∠OPO' = ∠OQO', let us join OP, O'P, OQ, O'Q and OO'. In ΔOPO' and ΔOQO', we have OP = OQ [Radii of the same circle] O'P= O'Q [Radii of the same circle] OO' = OO' [Common] ∴ Using the SSS criterion, ΔOPO'≌ ΔOQO' ∴ Their corresponding parts are equal. Thus, ∠OPO' = ∠OQO' 2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle. Solution: We have a circle with centre O. (Chord AB) || (Chord CD) and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm. Let ‘r’ be the radius of the circle. Let us draw OP ⊥ AB and OQ ⊥ CD. We join OA and OC. Let OQ = x cm ∴ OP = (6 ∠ x) cm [∵ PQ = 6 cm] ∵ The perpendicular from the centre of a circle to chord bisects the chord. 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre? REMEMBER The longer chord in a circle is nearer to the centre than the smaller chord. We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre. Let us draw OP ⊥ AB and join OA and OC. ∴ OP ⊥ AB ∴ P is the mid-point of AB. ⇒ AP = (1/2)AB = (1/2)(6 cm) = 3 cm Similarly, CQ = (1/2)CD = (1/2)(8 cm) = 4 cm Now in right ΔOPA, we have OA2 = OP2 + AP2 ⇒ r2 = 42 + 32 ⇒ r2 = 16 + 9 = 25 ⇒ r=√25 = 5 cm Note: r2 = 25 ⇒ r = ±5 But we reject r = ∠5, because distance cannot be negative. Again, in right ΔCQO, we have OC2 = OQ2 + CQ2 ⇒ r2 = OQ2 + 42 ⇒ OQ2 = r2 ∠ 42 ⇒ OQ2 = 52 ∠ 42 = 25 ∠ 16 = 9 [∵ r = 5 cm] ⇒ OQ =√9= 3 cm The distance of the other chord (CD) from the centre is 3 cm. Note: In case we take the two parallel chords on either side of the centre, then In ΔPOA, OA2 =OP2 + PA2r2 =42 + 32 = 52 ⇒ r = 5 cm In ΔQOC, OC2 =CQ2 + OQ2 ⇒ r2 =42 + OQ2 ⇒ OQ2 = 52 - 42 = 9 ⇒ OQ = 3 cm 4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. We have an ∠ABC such that the arms BA and BC on production, make two equal chords AD and CE. Let us join AC, DE and AE ∵ An exterior angle of a triangle is equal to the sum of interior opposite angles. ∴ In ΔBAE, we have Exterior ∠DAE = ∠ABC + ∠AEC …(1) ∵ The chord DE, subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle. ∴ ∠DAE = (1/2)∠DOE …(2) Similarly, ∠AEC =(1/2)∠AOC …(3) From (1), (2) and (3), we have (1/2)∠DOE = ∠ABC + (1/2)∠AOC ⇒ ∠ABC = (1/2)∠DOE - (1/2)∠AOC ⇒ ∠ABC = (1/2) [∠DOE - ∠AOC] ⇒ ∠ABC = (1/2) [(Angle subtended by the chord DE at the centre) ∠ (Angle subtended by the chord AC at the centre)] ⇒ ∠ABC =(1/2) [Difference of the angles subtended by the chords DE and AC] 5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals. Solution: We have a rhombus ABCD such that its diagonals AC and BD intersect at O. Taking AB as diameter, a circle is drawn. Let us draw PQ || AD and RS || AB, both passing through O. P, Q, R and S are the mid-points of DC, AB, AD and BC respectively. ∵ All the sides of a rhombus are equal. ∴ AB = DC ⇒ (1/2) AB =(1/2) DC ⇒ AQ = DP ⇒ BQ = DP ⇒ AQ = BQ [∵ Q is the mid-point of AB] ...(1) Similarly, RA = SB ⇒ RA = QO [∵ PQ is drawn parallel to AD] …(2) From (1) and (2), we have QA = QB = QO i.e. A circle drawn with Q as centre, will pass through A, B and O. Thus, the circle passes through the point of intersection (O) of the diagonals of the rhombus ABCD.