In Fig. 8, O is the center 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.

(CBSE 2016)

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:

OP² + PT² = OT²

(5)² + PT² = (13)²

_PT² = 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:

AE² + ET² = AT²

x² + (8)² = (12 – x)²

x² + 64 = 144 + x² – 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