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

Dear Student!

Here is the answer to your query.

Given, PQ is the chord of the circle and PT and QT are the tangents drawn at the end points of the chord PQ. PQ = 8cm and OP = 5cm.

OT ⊥ PQ,

∴ PR = RQ =  (Perpendicular from the centre of the circle to a chord bisect the chord)

In right  ΔOPR,

OP² = PR² + OR²

⇒ OR² = OP² – PR²

⇒ OR² = (5cm)² – (4cm²) = (25 – 16)cm²

⇒ OR² = 9cm²

⇒ OR = 3cm

In right ΔPTR,

PT² = TR² + PR²  ...(1)

We know that, the tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠OPT = 90º

In right ΔOPT,

OT² = PT² + OP²  ...(2)

From (1) and (2), we get

OT² = (TR² + PR²) + OP²

⇒ (TR + OR)² = (TR² + PR²) + OP²

⇒ TR² + OR² + 2 × TR × OR = TR² + PR² + OP²

⇒ 9 + 6 TR = 16 + 25

⇒ 6TR = 25 + 16 – 9 = 32

⇒ TR =

∴ PT² = TR² + PR²  (Using (1))

⇒ PT² = + (4cm)²

⇒ PT² =

⇒ PT

Thus, the length of tangent PT is .

Cheers!