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Question
In Figure 2, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.
Solution
PT = TQ (Tangents drawn from a common point to the circle are equal)
So, ΔTPQ is isosceles with PT = TQ and TO is thus the angle bisector of ∠PTQ.
So, OT ⊥⊥ PQ and thus, OT bisects PQ. (Perpendicular drawn from the center of the circle to the chord bisects the chord)
Thus, PR = RQ = 4 cm
Also, in ΔOPR,
OR =`sqrt("OP"^2-"PR"^2)`
`=sqrt(5^2 - 4^2)`
= 3 cm
In ΔTPR,
TR2 + PR2 = TP2
⇒ TR2 + 42 = TP2
⇒ TR2 + 16 = TP2
Let TP = x and TR = y
⇒ Y2 + 16 = X2 .....(1)
In Δ OPT,
TP2 + OP2 = OT2
⇒ TP2 + 52 = (TR + 3)2
⇒ X2 + 25 = (Y + 3)2.................(2)
Using (1) and (2) we get
y =`16/3`
putting this value of y in (1) we get
`(16/3)^2 + 16 = "x"^2`
⇒ `256/9 + 16 = "x"^2`
⇒ `(256+144)/9 ="x"^2`
⇒ `400/9 = "x"^2`
⇒ `"x" = 20/3`
Thus, the length of TP =`20/3`
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