Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers 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.
OA = OB = 5 cm
O'A = O'B = 3 cm
The diagram is shown below:
OO' will be the perpendicular bisector of chord AB
⇒ AC = CB
It is given that, OO' = 4 cm (The distance between their centres)
Let OC be x. Therefore, O'C will be 4 − x
In ΔOAC,
OA2 = AC2 + OC2
⇒ 52 = AC2 + x2
⇒ 25 – x2 = AC2 ..........(i)
Now, In ΔO'AC,
O'A2 = AC2 + O'C2
⇒ 32 = AC2 + (4 − x)2
⇒ 9 = AC2 + 16 + x2 − 8x
⇒ AC2 = − x2 − 7 + 8x ........(ii)
From (i) and (ii), we get
25 – x2 = − x2 − 7 + 8x
8x = 32
x = 4
Hence, O'C = 4 - 4 = 0 cm i.e. 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
Also, AC2 = 25 – x2
= 25 – 42
= 25 − 16
= 9
⇒ AC = 3 cm
⇒ Length of the common chord AB = 2 AC = (2 × 3) cm = 6 cm