Class IX Math NCERT Solution for Circles 10.4

EXERCISE 10.4 (Page 179)
Q1.  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 chrod.
Sol: 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 teh common chord.
          ∴ ∠OLP = ∠OLP = 90°
               PL = PQ
          Now in right ∠OLP,
               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 = 3cm
          PQ = PL + LQ = 3 cm + 3 cm = 6 cm
          Thus, the required length of the common chord = 6 cm.
Q2.  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.
Sol: 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 equidisitant from teh centre.]
          Now, in right ΔOME and right ΔONE,
               OM = ON                              [Proved]
               OE = OE                                 [Common]
          ∴ ΔOME ≌ ΔONE                   [RHS criterion]
               ME = NE                                [c.p.c.t]
          
          ⇒ 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
Q3.  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.
Sol: We have a circle with centre O and equal chords AB and CD are intersecting at E. OE is joined.
          A 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
Q4.  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).
Sol: 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 = MD                                        [Perpendicular from the centre to the chord 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
Q5.  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 hall 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?
Sol: 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 = 180°                                           [Linear pairs]
          ∴ ∠OPR = ∠OPM = 90°
          ⇒ OP⊥RM
          Now, in A RSP and A MSP,                                           [6 m each]
          RS = MS                                                                                [Common]
          SP = SP
          ∠RSP = ∠MSP
          ∴ΔRSP ≌ΔMSP                                                                   [SAS criterion]
          But ∠RSP + ∠MSP = 180°
          ⇒∠RPS = ∠MSP = 90°                                                     [Each]
          ∴ SP passes through O.
          Let OP = x m
          ∴ SP = (5 – x) m
          Now, in right ΔOPR,
               x2 + RP2 = 52
          In right ΔSPR,
               (5 – x)2 + RP2 = 62
          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
          
          Now, RP2 = 52 – x2 ⇒RP2 = 25 – (1.4)2
          ⇒ RP2 = 25 – 1.96 = 23.04 m
          
          ∴ RM = 2 RP = 2 × 4.8 m = 9.6 m
          Thus, distance between Reshma and Mandip is 9.6 m.
Q6.  A circular park of radius 20 m is situated in a colony. David are sitting at equal distance on its boundary each having a toy each other. Find the length of the string of each phone.
Sol: In the figure, let Ankur, Syed and David are sitting at A, S
                 AS = SD = AD
          i.e. DASD is an equilateral triangle.
          Let the length of each side of the equilateral triangle is 2x metres.
          Let us draw AM ⊥ SD.
          Since DASD is an equilateral,
          ∴ AM passes through O.
          

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