Question

Prove that the intercept of a tangent between a pair of parallel tangents to a circle subtend a right angle at the centre of the circle

Answer 1

Given: XY and X'Y' at are two parallel tangents to the circle wth centre O and AB is the tangent at the point C,which intersects XY at A and X'Y' at B.

To prove: ∠AOB = 90º.

Construction: Join OC.

Proof: In ΔOPA and ΔOCA

OP = OC (Radii )

AP = AC(Tangents from point A)

AO = AO (Common )

ΔOPA ≅ ΔOCA (By SSS criterion)

Therefore, ∠POA = ∠COA .... (1) (By C.P.C.T)

Similarly , ΔOQB ≅ ΔOCB ∠QOB = ∠COB .......(2)

POQ is a diameter of the circle. Hence, it is a straight line. Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°

From equations (1) and (2), it can be observed that

2∠COA + 2∠COB = 180°

∴ ∠COA + ∠COB = 90°

∴ ∠AOB = 90°.

Hence Proved....

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Answer 2

Given: XY and X

′

Y

′

are two parallel tangents to the circle wth centre O and AB is the tangent at the point C,which intersects XY at A and X

′

Y

′

at B.

To Prove:∠AOB=90

∘

Construction:Join OC

Proof:In △OPA and △OCA

OP=OC (radii)

AP=AC (Tangents from point A)

AO=AO (Common )

△OPA≅△OCA (By SSS criterion)

∴,∠POA=∠COA .... (1) (By C.P.C.T)

OQ=OC (radii)

BQ=BC (Tangents from point A)

BO=BO (Common )

△OQB≅△OCB

∠QOB=∠COB .......(2)

POQ is a diameter of the circle.

Hence, it is a straight line.

∴,∠POA+∠COA+∠COB+∠QOB=180

∘

From equations (1) and (2), it can be observed that

2∠COA+2∠COB=180

∘

∴∠COA+∠COB=90

∘

∴∠AOB=90

∘