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Question

Two circle of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. The length of their common chord is

A
12cm
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B
3cm
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C
6cm
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D
8cm
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Solution

The correct option is B 6cm
GivenA&Baretwointersectingcircles,thecomonchordofwhichisCD.CDintersectsABorextendedABatE.AB=4cm,radiusofthecirclewithcentreAisAC=5cmandradiusofthecirclewithcentreBisBC=3cm.TofindoutThelengthofthecommonchordCD.SolutionAEC=90osincethelinejoiningthecentresofoftwointersectingcirclesisperpendiculartotheircommonchord.Sowehavetworighttriangles,ΔACE&ΔBCEwithhypotenusesAC&BCrespectively.LetBE=xcm.ThenAE=(4+x)cmandBE=xcm.ByPythagorastheorem,wegetCE=AE2AC2=BC2BE252(4+x)2=32x2x=0.i.eB&Earesamepoint.AB=AE=4cm.SoCE=AE2AC2=52(4+0)2cm=3cm.NowCD=2CEsincethelinejoiningthecentresofoftwointersectingcirclesbisectsthecommonchord.ThelengthofthecommonchordCD=2×3cm=6cm.AnsOptionC.
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