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Maths / Circles / Relation of Chord Between The Point Of Contact and Tangent
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## Relation of Chord Between The Point Of Contact and Tangent

Relation of Chord Between The Point Of Contact and Tangent: From a point outside a circle two tangents are drawn. The angle of Chord joining the point of contact with the radius is half of the angle between the to tangents,

 Given: Two tangents BP and BQ are drawn to a circle with centre O from an external point TB.To Prove:  $\large \dpi{120} \large \angle PBQ=2\angle OPQ$.Proof:  Let    $\dpi{120} \large \angle PBQ=\theta$Now, We know that  BP = BQ. So, BPQ is an isosceles triangles.   $\dpi{120} \large \therefore \;\;\angle BPQ=\angle BQP=\frac{1}{2}(180^{\circ}-\theta )=90^{\circ}-\frac{1}{2}\theta$   $\dpi{120} \large \angle OPT=90^{\circ}$    [ Angle of the radius with the tangent]So,    $\dpi{120} \large \angle OPQ=\angle OPB-\angle BPQ=90^{\circ}-\left [ 90^{\circ}-\frac{1}{2}\theta \right ]$         $\dpi{120} \large =\frac{1}{2}\theta =\frac{1}{2}\angle PBQ$         This gives                 $\dpi{120} \large \angle PBQ=2\angle OPQ$

ILLUSTRATION: In the figure PA and PB are two tangents from point P to the circle prove that $\dpi{120} \large OA^2= OP X OT$

 Solution :  OP is the bisector of $\large \angle AOB$ and $\large \angle APB$Now $\large \Delta AOB$ is an isosceles as OA = OB   [ both radii of same circle]and OT is the angle bisectorTherefore OT $\large \perp$ ABIn $\large \Delta OAT$and $\large \Delta OAP$   $\large \angle O = \angle O$                     [Common] $\large \angle T = \angle A$                       [ Each $\large 90^0$ ] $\large \Delta OAT\sim \Delta OPA$     [ By AA Similarity Criteria]$\dpi{120} \large \Rightarrow \frac{OA}{OP}=\frac{AT}{PA}= \frac{OT}{OA}$$\dpi{120} \large \Rightarrow \frac{OA}{OP}=\frac{OT}{OA}$$\dpi{120} \large \Rightarrow \;\;OA^2= OP\; X \;OT$

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