Given−Oisthecentreofacircleofradius5cm.Tisapointoutsidethecircleatadistanceof13cmfromO.OTintersectsthecircleatEQT&PTaretangentsfromTtothecircleatQ&P.ABisthetangenttothecircleatEanditintersectQT&PTatA&Brespectively.Tofindout−ThelengthofAB.Solution−QT&PTaretangentsfromTtothecircleatQ&P.QT=PT.SoinΔOPT&ΔOQTwehaveQT=PT(proved),OTcommonside,andOP=OQ(radiiofthesamecircle.)∴ΔOPT&ΔOQTarecongruentbySSStest.∴∠PTO=∠QTO........(i)NowABisthetangenttothecircleatEandOEorOTisthelinejoiningthecentreOwiththepointofcontactE.∴OEorOT⊥AB⟹∠AET=90o=∠BET.....(ii)SoinΔAET&ΔBETwehave∠PTO=∠QTO....(from.(i))i.e∠ATO=∠BTO.∠AET=∠BET(fromii),andsideETiscommon.∴ByASAtestΔAET&ΔBETarecongruent.⟹AE=BE........(iii)Again∠OPT=90o(OPistheradiusdrawntothepointofcontactPofthetangentPT).....(iv)∴BetweenΔOPT&ΔABTwehave∠OPT=∠AET(fromii&ivbothare=90o),∠AET=∠BET(fromii).∴ΔOPT&ΔABTaresimilar.⟹PTET=OPAE=OP12AB⟺AB=2OPPT×ET..........(v)NowΔOPTisarightonewithhypotenuseOT,OP=OE=5cm,OT=13cm∴PT=√OT2−OP2=√132−52cm=12cmandET=OT−OE=(13−5)cm=8cm∴In(v)wehaveAB=2OPPT×ET.i.eAB=2×512×8 cm=203cm.∴AB=203cm.