Question

PQ is a chord of length 8 CM of a circle of radius 5 CM. The tangents at P and Q intersects at point T. Find the length of TP.

Answer 1

Answer.


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.