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Home / Geometry / Solution of exercise 8.4 (Circle) of class Nine and Ten ( 9 – 10)

Solution of exercise 8.4 (Circle) of class Nine and Ten ( 9 – 10)

Problem – 1. From some exterior point P of a circle with center O two tangents are drawn to the circle. Prove that OP is the perpendicular bisector of the chord of contact.

Particular Enunciation: Given that, O and P are the centre and any exterior point respectively of the circle ABC. Two tangents PA and PB are drawn to the circle. (A, B) and (O, P) are jointed. Let us prove that OP is perpendicular bisectors of the chord of contact AB.

Construction: let us join (O, A) and (O, B).

Proof: In the right angled triangles OPA and OPB we get,

AP = BP      [Length of the tangents from exterior point are equal]

OA= OB     [Radii of the same circle]

Since OA and OB are the radii passing through the points of contact,

∴ ∠PAO = 1right angle= ∠PBO

Hence the two triangles are congruent.

∴  ∠POA = ∠POB

Now in OAE and OBE we get OA=OB,

∠AOE= ∠BOE and OE is common side.

Hence the two triangles are congruent.

∴  AE= BE and ∠OEA =∠OEB, but these two are adjacent angles.

So, each them is one right angle.

∴  EO = AB. But EO and PO are the same line. Hence, OP is the perpendicular bisector of the chord of contact AB (Proved).

                                                

Problem – 2 .Given that O is the centre of the circle and the two tangents PA and PB touch the circle at the points A and B respectively. Prove that PO bisects  ∠APB.

General Enunciation: Given that O is the centre of the circle and the two tangents PA and PB touch the circle at the points A and B respectively. We have to Prove that PO bisects ∠BPA.

Particular Enunciation: Given that O is the centre of the circle and the two tangents PA and PB touch the circle at the points A and B respectively (P, O) are joined. Let us prove that PO bisects ∠BPA i.e. ∠OPA =∠OPB.

Construction: Let us join (O, A) and (O,B).

Proof: In OPA and OPB we get

AP =BP [Length of the tangents from exterior point are equal]

AO=BO [Radii of the same circle]                          

                And OP is the common side.

Hence the two triangles are congruent.

∴  ∠OPA= ∠OPB, i.e. OP is the bisector of ∠BPA (proved).

 

 

 

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