Question

In the given figure, 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.

Answer 1

Given : CD = 7.8 cm , PD = 5 cm and PB = 4 cm
We have related diagram :


i ) Here PC and PA are two secants of the circle , We know from Intersecting Secants Theorem " When two secant lines intersect each other outside a circle, the products of their segments are equal. " . So Here

PD × PC = PB × PA , Then

⇒ PD × ( PD + CD ) = PB × ( PB + AB ) , Now we substitute given values we get :

⇒ 5 × ( 5 + 7.8 ) = 4 × ( 4 + AB )

⇒ 5 × 12.8= 4 × ( 4 + AB )

⇒ 64= 4 × ( 4 + AB )

⇒ 16 = ( 4 + AB )

⇒ 16 = 4 + AB

⇒ AB = 12 cm ( Ans )

ii ) Here we have tangent = PT

We know from tangent-secant theorem " When a tangent and a secant are drawn from one single external point to a circle, square of the length of tangent segment must be equal to the product of lengths of whole secant segment and the exterior portion of secant segment." So here

We take Secant = PC and get :

$P{T}^2$ = PD × PC

⇒$P{T}^2$ = PD × ( PD + CD ) , Now we substitute given values we get :

⇒$P{T}^2$ = 5 × ( 5 + 7.8 )

⇒ $P{T}^2$ = 5 × 12.8

⇒ $P{T}^2$ = 64

⇒ $P{T}^2=\sqrt{64}$

$\Rightarrow PT=8$ cm ( Ans )