Question

ln Figure, PQ is a chord of length 8 cm of a circle of radius 5 cm and centre O. The tangents at P and Q intersect at point T. find the length of TP.

Answer 1

Given radius, OP = OQ = 5 cm

Length of the chord, PQ = 8 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 = 3 cm

In right ΔPTM,

PT² = TM² + PM² ......(1)

∠OPT = 90°  ......[Radius is perpendicular to the tangent at the 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 + QP²

⇒ OM² + 2 × TM × OM = PM² + OP² 

⇒ 9 + 6TM = 16 + 25

⇒ 6TM = 32

⇒ TM = `32/6 = 16/3`

Equation(1) becomes,

PT² = TM² + PM²

= `(16/3)^2 + 4^2`

= `(256/9) + 16`

= `(256 + 144)/9`

= `(400/9)`

= `(20/3)^2`

PT = `20/3`

This gives `("TP")/("PO") = ("RP")/("RO")`, i.e., `("TP")/5 = 4/3` or TP = `20/3` cm.