Prove that the centre lies on the bisector of the angle between the two tangents drawn from an external point to a circle
Answers
Step-by-step explanation:
Given: PT and TQ are two tangent drawn from an external point T to the circle.
To prove. \angle PTO = \angle QTO∠PTO=∠QTO where O is centre of circle.
Construction: Join OT.
We know that a tangent to circle is perpendicular to the radius at the point of contact.
therefore,
\angle{OPT} = \angle{OQT} = 90^{\circ}∠OPT=∠OQT=90
∘
In triangle OPT and OQT,
OT = OTOT=OT (Common)
OP = OQOP=OQ ( Radii of the circle)
\angle{OPT} = \angle{OQT}∠OPT=∠OQT (each is 90^{\circ}90
∘
)
So, Triangle OPT and OQT (RHS congruence criterion)
PT = TQPT=TQ and \angle{OTP} = \angle{OTQ}∠OTP=∠OTQ (CPCT)
PT = TQPT=TQ
Therfore
The lengths of the tangents drawn from an external point to a circle are equal.
\angle{OTP} = \angle{OTQ}∠OTP=∠OTQ
Therefore
Centre lies on the bisector of the angle between the two tangents.
So, option A is the answer.
solution
Step-by-step explanation:
Given: PT and TQ are two tangent drawn from an external point T to the circle.
To prove. \angle PTO = \angle QTO∠PTO=∠QTO where O is centre of circle.
Construction: Join OT.
We know that a tangent to circle is perpendicular to the radius at the point of contact.
therefore,
\angle{OPT} = \angle{OQT} = 90^{\circ}∠OPT=∠OQT=90
∘
In triangle OPT and OQT,
OT = OTOT=OT (Common)
OP = OQOP=OQ ( Radii of the circle)
\angle{OPT} = \angle{OQT}∠OPT=∠OQT (each is 90^{\circ}90
∘
)
So, Triangle OPT and OQT (RHS congruence criterion)
PT = TQPT=TQ and \angle{OTP} = \angle{OTQ}∠OTP=∠OTQ (CPCT)
PT = TQPT=TQ
Therfore
The lengths of the tangents drawn from an external point to a circle are equal.
\angle{OTP} = \angle{OTQ}∠OTP=∠OTQ
Therefore
Centre lies on the bisector of the angle between the two tangents.
So, option A is the answer.
solution
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