Question
Fri March 08, 2013 By:
 

2 circles of radii 5cm and 7 cm intersect at 2 points. THe distance between the centres of the 2 circles is 9 cm. Find the lenght of the common chord.

Expert Reply
Sat March 09, 2013
Answer : Given :2 circles of radii 5cm and 7 cm intersect at 2 points. The distance between the centres of the 2 circles is 9 cm.
To find : the lenght of the common chord.
 
Let the radius of the circle centered at O and O' be 5 cm and 7cm respectively.
OA = OB = 5 cm
O'A = O'B = 7 cm
OO' will be the perpendicular bisector of chord AB.
therefore AC = CB
It is given that, OO' = 9 cm
Let OC be x. Therefore, O'C will be 9 − x.
In triangle OAC,
OA= AC2 + OC2
=> 52 = AC2 + x2
=> 25 − x2 = AC2 ... (1)
In  triangle O'AC,
O'A2 = AC2 + O'C2
=> 72 = AC2 + (9 − x)2
=>49 = AC2 + 81 + x2 âˆ’ 18x
⇒ AC2 = − x2 âˆ’ 32 + 18x ... (2)
From equations (1) and (2), we obtain
25 − x= âˆ’ x2 âˆ’ 32 + 18x
18x = 57
x = 57/18 
=> x = 19 /6 = 3.17 cm
therefore OC = 19/6 cm and CO'= 9-x = 9-(19/6) = 35/6 cm
putting the value in eq 1 
AC2 = 25 - x2
=> AC2 = 25 - (19/6)
=> AC2 = (900 - 361 ) / 36
=> AC2 = (900 - 361 ) / 36
=> AC2 = 539 /36 = 14.97 
=> AC = (14.97)1/2 = 3.87 cm (approx)
The lenght of chord = 2AC = 2* 3.87
= 7.74 cm (approx) Answer

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