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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=2angle OPQ.

Proof:  Let    large angle PBQ=theta

Now, We know that  BP = BQ. So, BPQ is an isosceles triangles.

   large therefore ;;angle BPQ=angle BQP=frac{1}{2}(180^{circ}-theta )=90^{circ}-frac{1}{2}theta

   large angle OPT=90^{circ}    [ Angle of the radius with the tangent]

So,    large angle OPQ=angle OPB-angle BPQ=90^{circ}-left [ 90^{circ}-frac{1}{2}theta right ]         large =frac{1}{2}theta =frac{1}{2}angle PBQ        

 This gives                 large angle PBQ=2angle OPQ

ILLUSTRATION: In the figure PA and PB are two tangents from point P to the circle prove that 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 bisector

Therefore OT large perp AB

In large Delta OATand large Delta OAP

   large angle O = angle O                     [Common]

 large angle T = angle A                       [ Each large 90^0 ]

 large Delta OATsim Delta OPA     [ By AA Similarity Criteria]

large Rightarrow frac{OA}{OP}=frac{AT}{PA}= frac{OT}{OA}

large Rightarrow frac{OA}{OP}=frac{OT}{OA}

large Rightarrow ;;OA^2= OP; X ;OT

 

 


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