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.
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
⇒25−x2=AC2...(1)
In ΔO′AC,
O′A2=AC2+O′C2
⇒32=AC2+(4−x)2
⇒9=AC2+16+x2−8x
⇒AC2=−x2−7+8x...(2)
From equations (1) and (2), we obtain 25−x2=−x2−7+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=25−x2=25−42=25−16=9
∴AC=3cm
Length of the common chord AB = 2 AC = 2×3 cm = 6 cm