Relation of Chord Between The Point Of Contact and Tangent


 
 
Concept Explanation
 

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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