Question

Two circles of radii 5 cm5 cm. Find the length (in metre) of the common chord.

Answer 1

Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'
Hence OA = OB = 5 cm
O'A = O'B = 3 cm
OO' is the perpendicular bisector of chord AB.
Therefore, AC = BC
Given OO' = 4 cm
Let OC = x
Hence O'C = 4 − x
In right angled ΔOAC, by Pythagoras theorem

$O{A}^2=O{C}^2+A{C}^2$

⇒$5^2=x^2+A{C}^2$

⇒$A{C}^2=25-x^2$ (1)

In right angled $\Delta$O'AC, by Pythagoras theorem

$O^{\prime }{A}^2=A{C}^2+O^{\prime }{C}^2$
⇒$3^2=A{C}^2+(4-x{)}^2$

⇒$9=A{C}^2+16+x^2-8x$

⇒$A{C}^2=8x-x^2-7$ (2)

From (1) and (2), we get

$25-x^2=8x-x^2-7$

$8x=32$
Therefore, x = 4
Hence the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.

$A{C}^2=25-x^2$

= $25-4^2$

= 25 − 16 = 9

Therefore, AC = 3 cm
Length of the common chord, AB = 2AC = 6 cm