# In Fig. 2,

Join the point T with O.

By theorem, we know that the lengths of tangents from the external point is equal.
So, TP = TQ
In ΔTPQ,

TP = TQ

As two sides are equal, so the ΔTPQ is an isosceles triangle.

Here, OT is the bisector of ∠PTQ,

So, OT ⊥ PQ

∴ PR = RQ (perpendicular from centre to the chord bisects the chord)

So,

PR = QR = ½ PQ

= ½ (8)

= 4 cm

In ∆ OPR

OP2 = OR2 + PR2

52 = OR2 + 42
⇒ 25 = OR2 + 16
⇒  9 = OR2

⇒ OR = 3 cm

In ∆ OPT

OT2 = OP2 + PT2

(TR + 3)2 = 52 + PT2  …(1)

In ∆ PRT

PT2 = TR2 + RP2

PT2 = TR2 + 42 …(2)

Put the value of PT2 in (1),

⇒ (TR + 3)2 = 52 + TR2 + 42

⇒ TR2 + 9 + 6TR = 25 + TR2 + 16

⇒ 6TR + 9 = 25 + 16

⇒ 6TR  = 25 + 16 - 9

⇒ 6TR  = 32

From (2),

PT2 = (16/3)2 + 16

So, TP = 20/3 cm

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