Question

In Fig. O is the centre of a circle of radius 5 cm T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.

Answer 1

Given OT = 13 cm and OP = 5cm

If we draw a line from the centre to the tangent of the circle it is always perpendicular to the tangent

i.e . OP ⊥ PT.

In right angled Δ OPT,

OT² = OP² + PT²

[ by Pythagoras theorem:

(hypotenuse)² = (base)² + (perpendicular)²]

⇒ PT² = (13)² − (5)² = 169 − 25 = 144

⇒ PT = 12 cm

Since, the length of pair of tangents from an external point T is equal.

∴ QT = 12 cm

Now. TA = PT - PA …………(i)

⇒ TA = 12 - PA

And TB = QT - QB ……..(ii)

⇒ TB = 12 - QB

Again, using the property, length of pair of tangents from an external point is equal.

∴ PA = AE and QB = EB ……..(iii)

OT = 13 cm (Given)

∴ ET = OT - OE [ ∴ OE = 5cm = radius]

⇒ ET = 13 - 5

⇒ ET = 8cm

Since AB is a tangent OE is the radius

∴ OE ⊥ AB

⇒ OEA = 90^∘ [ linear pair]

∠AET + ∠OEA = 180^∘

∴∠AET = 180^∘ −∠OEA 

⇒∠AET = 90^∘

Now, in right angled ΔAET,

(AT)² = (AE)² + (ET)² [by Pythagoras theorem]

⇒ (PT − PA)² = (AE)² + (8)²

⇒ (12−PA)² = (PA)² + (8)² (from eq(iii)]

⇒ 144 + (PA)² − 24PA = (PA)² + 64

⇒ 24PA = 80

⇒ PA = 10/3 cm

⇒ AE = 10/3cm [from(iii)]

BE = 10/3 cm

AB = AE + EB

= 10/3 + 10/3

= 20/3 cm

Hence the required length AB is 20/3 cm.

Answer 2

Thus   AB = 2 * x = 2 *3.3

= 6.6 cm.

Comments

Why is AB equal to 2AE

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see the above diagram

Answer 3

Thus, AB=6.6cm