Question

O is the centre of a circle of radius 6 cm. P is a

point such that OP = 10 cm and OP intersects the

circle at T and PC, PD are two tangents drawn to

the circle. If AB is the tangent to the circle at T,

find the length AB​

Answer 1

Answer:

Clearly ∠OPT=90

o

Applying Pythagoras in △OPT, we have

OT

2

=OP

2

+PT

2

⇒13

2

=5

2

+PT

2

⇒PT

2

=169−25=144

⇒PT=12 cm

Since lengths of tangents drawn from a point to a circle are equal. Therefore,

AP=AE=x(say)

⇒AT=PT−AP=(12−x)cm

Since AB is the tangent to the circleE. Therefore, OE⊥AB

⇒∠OEA=90

o

⇒∠AET=90

o

⇒AT

2

=AE

2

+ET

2

[Applying Pythagoras Theorem in △AET]

⇒(12−x)

2

=x

2

+(13−5)

2

⇒144−24x+x

2

=x

2

+64

⇒24x=80

⇒x=

3

10

cm

Similarly, BE=

3

10

cm

∴AB=AE+BE=(

3

10

+

3

10

)cm=

3

20

cm

hope it's correct and help full