Question

Two circles with centres $A$ and $B$ of radii $3cm$ and $4cm$, respectively intersect at two points $C$ and $D$ such that $AC$ and $BC$ are tangents to the two circles. Find the 10 times length of the common chord $CD$

• A. 48
• B. 58
• C. 56
• D. 54

Answer 1

The correct option is A 48

$GivenOand{O}^{\prime }aretwocirclesitersectingatP\&Qrespectively.OP\&O^{\prime }Paretangentstothecircleswithcenters{O}^{\prime }\&Orespectively.OP=OQ=3cmand{O}^{\prime }P=O^{\prime }Q=4cm.Tofindout-ThelengthofthecommonchordPQ.Solution-OP\&O^{\prime }ParetangentstothecirclesatthepointofcontactofthetangetsPandalsotheyarelinesjoiningthecentrestothepointofcontact.\therefore \angle OP{O}^{\prime }={90}^o.i.e\Delta OP{O}^{\prime }isarightonewithO{O}^{\prime }ashypotenuse.(Similarly\Delta OQ{O}^{\prime }isarightonewithO{O}^{\prime }ashypotenuse.)⟹{O^{\prime }O}^2={OP}^2+{O^{\prime }P}^2⟹O^{\prime }O=\sqrt{3^2+4^2}cm=5cm..........\left(i\right)Nowbetween\Delta OQ{O}^{\prime }\&\Delta OP{O}^{\prime }hypO{O}^{\prime }iscommon,side{O}^{\prime }P=O^{\prime }Q\Delta OQ{O}^{\prime }\&\Delta OP{O}^{\prime }arecongruent.⟹\angle POM=\angle QOM.Buttheyarelinearpair.\therefore \angle POM=\angle QOM={90}^o⟹\Delta POM\&\Delta QOMarerighttriangleswithOP\&OQashypotenuses.Sobetween\Delta POM=\Delta QOMwehavehypotenusesOP=OQandsideOP=OQ\left(radiiofthesamecircle\right)\therefore \Delta POMiscongruentto\Delta QOM⟹\angle OMP=\angle OMQ.Buttheyareadjacentangles.\therefore \angle OMP=OMQ={90}^o.......\left(ii\right)andPM=QM⟹2PM=PQ....\left(iii\right)Nowbetween\Delta O{O}^{\prime }P\&\Delta OPMwehave\angle OMP=\angle O{O}^{\prime }P={90}^o\left(fromii\right),\angle POMcommon.So\Delta O{O}^{\prime }P\&\Delta OPMaresimilar.\therefore \frac{PM}{3}=\frac{O^{\prime }P}{O{O}^{\prime }}⟹PM=3\times \frac{4}{5}cm=2.4cm.\left(fromi\&iii\right)i.ePQ=2PM=2\times 2.4cm=4.8cmAns-4.8cm.$