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. - Mathematics

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Sum

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.

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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`

Concept: Construction of Tangents to a Circle
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