Question

In Fig. 9.20. O is the centre of a circle of radius 5 cm, T is a point such that OT $=$ 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E and it intersects the tangents PT and QT at A and B, find the length of AB.

• A. $\frac{10}{3}$ cm
• B. $\frac{20}{3}$ cm
• C. $10$ cm
• D. $15$ cm

Answer 1

Answer: (B)

$Given-Oisthecentreofacircleofradius5cm.Tisapointoutsidethecircleatadistanceof13cmfromO.OTintersectsthecircleatEQT\&PTaretangentsfromTtothecircleatQ\&P.ABisthetangenttothecircleatEanditintersectQT\&PTatA\&Brespectively.Tofindout-ThelengthofAB.Solution-QT\&PTaretangentsfromTtothecircleatQ\&P.QT=PT.Soin\Delta OPT\&\Delta OQTwehaveQT=PT\left(proved\right),OTcommonside,andOP=OQ(radiiofthesamecircle.)\therefore \Delta OPT\&\Delta OQTarecongruentbySSStest.\therefore \angle PTO=\angle QTO........\left(i\right)NowABisthetangenttothecircleatEandOEorOTisthelinejoiningthecentreOwiththepointofcontactE.\therefore OEorOT\perp AB⟹\angle AET={90}^o=\angle BET.....\left(ii\right)Soin\Delta AET\&\Delta BETwehave\angle PTO=\angle QTO....(from.(i\left)\right)i.e\angle ATO=\angle BTO.\angle AET=\angle BET\left(fromii\right),andsideETiscommon.\therefore ByASAtest\Delta AET\&\Delta BETarecongruent.⟹AE=BE........\left(iii\right)Again\angle OPT={90}^o\left(OPistheradiusdrawntothepointofcontactPofthetangentPT\right).....\left(iv\right)\therefore Between\Delta OPT\&\Delta ABTwehave\angle OPT=\angle AET(fromii\&ivbothare={90}^o),\angle AET=\angle BET\left(fromii\right).\therefore \Delta OPT\&\Delta ABTaresimilar.⟹\frac{PT}{ET}=\frac{OP}{AE}=\frac{OP}{\frac{1}{2}AB}⟺AB=\frac{2OP}{PT}\times ET..........\left(v\right)Now\Delta OPTisarightonewithhypotenuseOT,OP=OE=5cm,OT=13cm\therefore PT=\sqrt{{OT}^2-{OP}^2}=\sqrt{{13}^2-5^2}cm=12cmandET=OT-OE=(13-5)cm=8cm\therefore In\left(v\right)wehaveAB=\frac{2OP}{PT}\times ET.i.eAB=\frac{2\times 5}{12}\times 8cm=\frac{20}{3}cm.\therefore AB=\frac{20}{3}cm.$