Answer :

In the given figure, PQ = 16 cm

PO = 10 cm (radius)

**⸫** TP = TQ [Tangents drawn from an external point to the same circle are equal]

ΔTPQ is an isosceles triangle

**⸫** OT is the perpendicular bisector of PQ

**⸫** PR = 8 cm

In ΔTPR,

TP^{2} = TR^{2} + PR^{2} [Pythagoras theorem]

TP^{2} = TR^{2} + 8^{2} …. (i)

∠TPO = 90° [A tangent is perpendicular to the radius at the point of tangency]

In ΔTPO,

TO^{2} = TP^{2} + PO^{2} [Pythagoras theorem]

TO^{2} = TP^{2} + 10^{2}

TP^{2} = TO^{2} – 100 …. (ii)

In ΔOPR,

OR^{2} = OP^{2} – PR^{2} [Pythagoras theorem]

OR^{2} = 10^{2} – 8^{2}

**⸫** OR = √36 = 6 cm

Equation eq (i) and eq (ii),

TR^{2} + 64 = TO^{2} – 100

TR^{2} + 64 = (TR + OR)^{2} – 100 [From the figure, (TR + OR) = TO]

TR^{2} + 64 = TR^{2} + 36 + 12TR – 100

12TR = 128

**⸫** TR =

Substituting the value of TR in eq (i), we get

TP^{2} =

TP^{2} =

**⸫** TP =

**⸫** TP = **13.33** **cm**

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