PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at point T find length of TP
intersect at a point T. Find the length TP.
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Answer:
Joint OT.
Let it meet PQ at the point R.
Then ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.
[∵TP=TQ= Tangents from T upon the circle]
∴OT⊥PQ
∴OT bisects PQ.
PR=RQ=4 cm
Now,
OR=
OP
2
−PR
2
=
5
2
−4
2
=3 cm
Now,
∠TPR+∠RPO=90
∘
(∵TPO=90
∘
)
=∠TPR+∠PTR(∵TRP=90
∘
)
∴∠RPO=∠PTR
∴ Right triangle TRP is similar to the right triangle
PRO. [By A-A Rule of similar triangles]
∴
PO
TP
=
RO
RP
⇒
5
TP
=
3
4
⇒TP=
3
20
cm.
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