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Question 1
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

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Solution


Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.
Also, OA = OB = 5
cm
O'A = O'B = 3 cm
OO' will be the perpendicular bisector of chord AB.
AC=CB
It is given that, OO' = 4 cm
Let OC be x.
Therefore, O'C will be 4 - x.
In ΔOAC,
OA2=AC2+OC2[using pythagoras theorem]
52=AC2+x2
25x2=AC2...(1)
In ΔOAC,
OA2=AC2+OC2
32=AC2+(4x)2
9=AC2+16+x28x
AC2=x27+8x...(2)
From equations (1) and (2), we obtain 25x2=x27+8x
8x = 32
x = 4
Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.

AC2=25x2=2542=2516=9
AC=3cm
Length of the common chord AB = 2 AC = 2×3 cm = 6 cm


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