Answer :

We know that, tangent of the circle is perpendicular to the radius through the point of contact

∴ ∠ OPT = 90^{o}

Also, it is given in the question that:

OP = 5 cm

OT = 13 cm

Now, in right triangle OPT by using the Pythagoras theorem we have:

OP^{2} + PT^{2} = OT^{2}

(5)^{2} + PT^{2} = (13)^{2}

PT^{2} = 144

PT = 12 cm

As radius of the circle are equal to each other

∴ OP = OQ = OE = 5 cm

Also, ET = OT – OE

= 13 – 5

= 8 cm

Let us now assume PA = x cm

∴ AT = (12 – x) cm

We know that, tangents of a circle drawn from an external point are equal

∴ PA = AE = x cm

Now, in right triangle Aet by using the Pythagoras theorem we have:

AE^{2} + ET^{2} = AT^{2}

x^{2} + (8)^{2} = (12 – x)^{2}

x^{2} + 64 = 144 + x^{2} – 24x

24x = 80

∴

Now, AB = AE + EB

= AE + AE

= 2 × AE

= 2x

Putting the value of x, we get:

= 6.67 cm

Hence, the length of the AB will be 6.67 cm

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