Question

In the figure 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: 

1) AB.

2) the length of tangent PT.

Answer 1

Theorem used: Product of the 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.

1) As chord CD and tangent at point T intersect each other at P,

`PC xx PD = PT^2` ...(i)

AB is the diameter and tangent at point T intersect each other at P,

`PA xx PB = PT^2`  ...(ii)

From (i) and (ii), PC × PD = PA × PB … (iii)

Given: PD = 5cm, CD = 7.8 cm

PA = PB + AB = 4 + AB, and PC = PD + CD = 12.8 cm

Subs. these values in (iii),

12.8 × 5 = (4 + AB) × 4

`=> 4  + AB = (12.8 xx 5)/4`

`=> 4 + AB = 16`

`=> AB = 12cm`

2) `PC xx PD =  PT^2`

`=> PT^2 = 12.8 xx 5 =64`

`=> PT = 8 cm`

Thus, the length of the tangent is 8 cm