Question

hey... it's a tough question plz answer... in fig. o is the center of a circle of radius 5cm. t is a point such that o t equal 13 cm and ot intersect circle at e if ab is tangent to circle at e find the length of a b TP and tq are tangents to the circle.

Answer 1

AB, TQ and TP are tangents to the given circle.

OQ and PO are the radii.

OQ = PO = 5 cm

OT = 13 cm

By theorem, Angle OQT = Angle OPT = 90°

By pythagoras theorem,

TO^2 = TQ^2 + OQ^2

13^2 = TQ^2 + 5^2

169 - 25 = TQ^2

TQ^2 = 144

TQ = 12 cm

Tangents are equal.

TP = TQ = 12cm

According to the figure,

PA = AE = EB = EQ

TQ = TP = 12cm

PA + AT = TP

AT = 12 - PA

AT = 12 - AE [ PA = AE ]

AB is the tangent to the circle,

OEA = OEB = 90°

Similarly, AET = TEB = 90°

TE = TO - OE

TE = 13 - 5 = 8 cm

By pythagoras theorem,

AT^2 = AE^2 + TE^2

( 12 - AE ) ^2 = AE^2 + 8^2

144 + AE^2 - 24 AE - AE^2 = 64

144 - 24 AE - 64 = 0

80 - 24 AE = 0

24 AE = 80

AE = 80/24

AE = 10/3

AB = AE + EB

AB =( 10/3) + (10/3) = 20/3 cm

Answer 2

your answer will be 20/3 cm I hope it help you