Question

in the figure ab is a chord of length 8 cm of a circle of radius 5 cm the tangent to the circle at A and B intersect at the P find the length of the AP

Answer 1

Answer:

Step-by-step explanation:

Given radius, OP = OQ = 5 cm

Length of chord, PQ = 4 cm

OT ⊥ PQ,

∴ PM = MQ = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]

In right ΔOPM,

OP2 = PM2 + OM2

⇒ 52 = 42 + OM2

⇒ OM2 = 25 – 16 = 9

Hence OM = 3cm

In right ΔPTM,

PT2 = TM2 + PM2 → (1)

∠OPT = 90º [Radius is perpendicular to tangent at point of contact]

In right ΔOPT,

OT2 = PT2 + OP2 → (2)

From equations (1) and (2), we get

OT2 = (TM2 + PM2) + OP2

⇒ (TM + OM)2 = (TM2 + PM2) + OP2

⇒ TM2 + OM2 + 2 × TM × OM = TM2 + PM2 + OP2

⇒ OM2 + 2 × TM × OM = PM2 + OP2

⇒ 32 + 2 × TM × 3 = 42 + 52

⇒ 9 + 6TM = 16 + 25

⇒ 6TM = 32

⇒ TM = 32/6 = 16/3

Equation (1) becomes,

PT2 = TM2 + PM2  

        = (16/3)2 + 42

        = (256/9) + 16 = (256 + 144)/9

        = (400/9) = (20/3)2

Hence PT = 20/3

Thus, the length of tangent PT is (20/3) cm