Question

PQ is a chord of length 8 cm of a circle of radius 5 cm. the tangets at point of contact P & Q intersect at point T. find the length of TP

Answer 1

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.