# PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP. Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T

To Find : Length of TP

Construction : Join OQ Now in OPT and OQT

OP = OQ [radii of same circle]

PT = PQ

[tangents drawn from an external point to a circle are equal]

OT = OT [Common]

OPT OQT [By Side - Side - Side Criterion]

POT = OQT

[Corresponding parts of congruent triangles are congruent]

or POR = OQR

Now in OPR and OQR

OP = OQ [radii of same circle]

OR = OR [Common]

POR = OQR [Proved Above]

OPR OQT [By Side - Angle - Side Criterion]

ORP = ORQ

[Corresponding parts of congruent triangles are congruent]

Now,

ORP + ORQ = 180° [Linear Pair]

ORP + ORP = 180°

ORP = 90°

OR PQ

RT PQ

As OR PQ and Perpendicular from center to a chord bisects the chord we have In right - angled OPR,

By Pythagoras Theorem

[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]

(OP)2 = (OR)2 + (PR)2

(3)2 = (OR)2 + (2.4)2

9 = (OR)2 + 5.76

(OR)2= 3.24

OR = 1.8 cm

Now,

In right angled TPR,

By Pythagoras Theorem

(PT)2 = (PR)2 + (TR)2 …

Also, OP OT

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

In right angled OPT, By Pythagoras Theorem

(PT)2 + (OP)2 = (OT)2

(PR)2 + (TR)2 + (OP)2= (TR + OR)2 …[From 1]

(2.4)2 + (TR)2 + (3)2 = (TR + 1.8)2

4.76 + (TR)2 + 9 = (TR)2 + 2(1.8)TR + (1.8)2

13.76 = 3.6TR + 3.24

3.6TR = 10.52

TR = 2.9 cm [Appx]

Using this in 

PT2 = (2.4)2 + (2.9)2

PT2 = 4.76 + 8.41

PT2 = 13.17

PT = 3.63 cm [Appx]

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