Question

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 lengths of TP and TQ.

Answer 1

Given,
Radius, OP = 5 cm
PQ = 8 cm,
TP = TQ                               (Tangents drawn from a common point to a circle are equal)
Δ PTQ is isosceles. 
Let PQ and OT intersect at point M.
TO is the angle bisector of ∠PTQ.
So, ∠PMT = 90° ⇒ PM = QM = 4 cm               ....  Perpendicular drawn from the center of the circle to the chord bisects the chord)

Given,
Radius, OP = 5 cm
PQ = 8 cm,
TP = TQ                               (Tangents drawn from a common point to a circle are equal)
Δ PTQ is isosceles. 
Let PQ and OT intersect at point M.
TO is the angle bisector of ∠PTQ.
So, ∠PMT = 90° ⇒ PM = QM = 4 cm               ....  Perpendicular drawn from the center of the circle to the chord bisects the chord)

In Δ PMO, 

OP² = OM² + PM²

⇒ 5² = OM² + 4²

⇒ OM = 3 cm

Let PT = x and TM = y

In Δ PMT, 

PT² = TM² + PM²

x² = y² + 16          .....(i)

In Δ TPO,

OT² = PT² + OP² 

(y + 3)² = x² + 5² 

y² + 9 + 6y = x² + 25      ....(ii)

From (i) and (ii), we get

y² + 9 + 6y = y² + 16 + 25

6y = 32

y = `16/3` cm

substituting y = `16/3`in equation (i),

`"x"^2 = (16/3)^2 + 16`

`x^2 = 400/9`

`x = 20/3` cm

Therefore, PT = `20/3` cm