1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. We have two intersecting circles with centres at O and O' respectively. Let PQ be the common chord. ∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord. ∴ ∠OLP = ∠OLP = 90º PL = PQ Now in right ΔOLP, PL2 + OL2 = OP2 ⇒ PL2 + (4 - x)2 = 52 ⇒ PL2 = 52 - (4 - x)2 ⇒ PL2 = 25 - 16 - x2 + 8x ⇒ PL2 = 9 - x2 + 8x ...(1) Again, in ΔO'LP, PL2 = 32 - x2 = 9 - x2 ...(2) From (1) and (2), we have 9 - x2 + 8x = 9 - x2 ⇒ 8x = 0 ⇒ x= 0 ⇒ L and O' coincide. ∴ PQ is a diameter of the smaller circle. ⇒ PL = 3 cm But PL = LQ ∴ LQ = 3 cm ∴ PQ = PL + LQ = 3 cm + 3 cm = 6 cm Thus, the required length of the common chord = 6 cm. 2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. We have a circle with centre O. Equal chords AB and CD intersect at E. To prove that AE = DE and CE = BE, draw OM ⊥ AB and ON ⊥ CD. Since AB = CD [Given] ∴ OM = ON [Equal chords are equidistant from the centre] Now, in right ΔOME and right ΔONE, OM = ON [Proved] OE = OE [Common] ∴ ΔOME ≌ ΔONE [RHS criterion] ⇒ ME = NE [c.p.c.t.] ⇒ (1/2) AE = (1/2) DE [Perpendiculars from the centre bisect the chord.] ⇒ AE = DE ...(1) Since AB = CD (Given) ∴ AB - AE = CD - DE ⇒ CE = BE ...(2) From (1) and (2), we have AE = DE and CE = BE 3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. We have a circle with centre O and equal chords AB and CD are intersecting at E. OE is joined. To prove that ∠1 = ∠2, let us draw OM ⊥ AB and ON ⊥ CD. In right ΔOME and right ΔONE, OM = ON [Equal chords are equidistant from the centre.] OE = OE [Common] ∴ ΔOME ≌ΔONE [RHS criterion] ⇒ Their corresponding parts are equal. ∴ ∠OEM = ∠OEN or ∠OEA = ∠OED 4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see figure). We have two circles with the common centre O. A line ‘l’ intersects the outer circle at A and D and the inner circle at B and C. To prove that AB = CD, let us draw OM ⊥ ℓ.For the outer circle, ∵ OM ⊥ ℓ [Construction] ∴ AM = M D [∵ Perpendicular from the centre bisects the chord] ...(1) For the inner circle, ∵ OM ⊥ ℓ [Construction] ∴ BM = MC [Perpendicular from the centre to the chord bisects the chord] ...(2) Subtracting (2) from (1), we have AM - BM = MD - MC ⇒ AB = CD 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip? Let the three girls Reshma, Salma and Mandip are positioned at R, S and M, respectively on the circle of radius 5 m.∵ RS = SM = 6 m [Given] ∵ Equal chords of a circle subtend equal angles at the centre. ∴ ∠1 = ∠2 In ΔPOR and ΔPOM, OP = OP [Common] OR = OM [Radii of the same circle] ∠1= ∠2 [Proved] ∴ ΔPOR ≌ ΔPOM [SAS criterion] ⇒ Their corresponding parts are equal. ∴ PR = PM and ∠OPR = ∠OPM ∵ ∠OPR + ∠OPM = 180º [Linear pairs] ∴ ∠OPR = ∠OPM = 90º ⇒ OP ⊥ RM Now, in ΔRSP and ΔMSP, RS = MS (6 m each) SP = SP (Common) ∠RSP = ∠MSP ∴ ΔRSP ≌ ΔMSP [SAS criterion] ⇒ RP = MP and ∠RPS = ∠MPS [c.p.c.t] But ∠RPS + ∠MPS = 180° ⇒ ∠RPS = ∠MPS = 90º (Each) ∴ SP passes through O. Let OP = x m ∴ SP = (5 - x) m Now, in right ΔOPR, x2 + RP2 = 52 …(1) In right ΔSPR, (5 - x)2 + RP2 = 62 …(2) From (1), RP2 = 52 - x2 From (2), RP2 = 62 - (5 - x)2 ∴ 52 - x2 = 62 - (5 - x)2 ⇒ 25 - x2 = 36 - [25 - 10x + x2] ⇒ 25 - x2 - 36 + 25 - 10x + x2 = 0 ⇒ -10x + 14 = 0 ⇒ 10x = 14 ⇒ x= (14/10) = 1.4 Now, RP2 = 52 - x2 ⇒ RP2 = 25 - (1.4)2 ⇒ RP2 = 25 - 1.96 = 23.04 m ∴ RP =√(23.04) = 4.8 m ∴ RM = 2 RP = 2 x 4.8 m = 9.6 m Thus, distance between Reshma and Mandip is 9.6 m. 6. A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. In the figure, let Ankur, Syed and David are sitting at A, S and D respectively such that AS = SD = ADi.e.ΔASD is an equilateral triangle. Let the length of each side of the equilateral triangle is 2x metres. Let us draw AM ⊥ SD. Since ΔASD is an equilateral, ∴ AM passes through O. ⇒ SM = (1/2) SD = (1/2) (2x) ⇒ SM = x Now, in right ΔASM, AM2 + SM2 = AS2 ⇒ AM2 = AS2 - SM2 = (2x)2 -x2 = 4x2 - x2 = 3x2 ⇒ AM = 3x Now, OM = AM - OA = ( 3x - 20) m Again, in right ΔOBM, we have OS2 = SM2 + OM2 ⇒ 202 = x2 + (3x - 20)2 ⇒ 400 = x2 + 3x2 - 403 x + 400 ⇒ 4x2 - 403 x= 0 ⇒ 4x (x - 103) = 0 ⇒ x = 0 or x = 103 But x = 0 is not required. ∴ x= 103 m Now, SD = 2x m = 2 x 103 m Thus, the length of the string of each phone = 2 x 103 m.