In the figure (ii) given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find :

(i) AB

(ii) the length of tangent 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.

∴ TP² = PC × PD

From figure,

PC = PD + CD
= 5 + 7.8
= 12.8 cm.

⇒ TP² = 12.8 × 5
⇒ TP² = 64
⇒ TP = $\sqrt{64}$
⇒ TP = 8 cm.

Similarly,

⇒ TP² = AP × BP
⇒ 8² = AP × 4
⇒ 64 = 4AP
⇒ AP = $\dfrac{64}{4}$
⇒ AP = 16 cm.

(i) From figure,

AB = AP - BP = 16 - 4 = 12 cm.

Hence, the length of AB = 12 cm.

(ii) The length of tangent PT = 8 cm.