Question

PQ is a chord of length 8 cm of a circle of radius 5 cm the tangent at P and Q intersect at a point P find the length TP

Answer 1

Answer. Given radius, OP = OQ = 5 cmLength 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 = 16/3Equation (1) becomes,  PT2 = TM2 + PM2          = (16/3)2 + 42          = (256/9) + 16 = (256 + 144)/9          = (400/9) = (20/3)2Hence PT = 20/3 Thus, the length of tangent PT is (20/3) cm.