PAB is a secant and PT is tangent to a circle. If

(i) PT = 8 cm and PA = 5 cm, find the length of AB.

(ii) PA = 4.5 cm and AB = 13.5 cm, find the length of PT.

We know that,

If a chord and a tangent intersect externally, then the product of lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ PT² = PA × PB

(i) Putting values in above equation:

⇒ 8² = 5 × PB
⇒ 8² = 5PB
⇒ PB = $\dfrac{64}{5}$
⇒ PB = 12.8 cm.

AB = PB - PA = 12.8 - 5 = 7.8 cm.

Hence, the length of AB = 7.8 cm.

(ii) We know,

PB = AB + PA = 13.5 + 4.5 = 18 cm.

PT² = PA × PB
⇒ PT² = 4.5 × 18
⇒ PT² = 81
⇒ PT = $\sqrt{81}$
⇒ PT = 9 cm.

Hence, the length of PT = 9 cm.