Question

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.

• A. 1 cm
• B. 2 cm
• C. 6 cm
• D. 3 cm

Answer 1

The correct option is C 6 cm


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.
$\therefore AC=CB$
It is given that, OO' = 4 cm

Let OC be x.
Therefore, O'C will be 4 - x.

In $\Delta OAC$,
$O{A}^2=A{C}^2+O{C}^2$
[using pythagoras theorem]

$\Rightarrow 5^2=A{C}^2+x^2$
$\Rightarrow 25-x^2=A{C}^2...\left(1\right)$

In $\Delta O^{\prime }AC,$
$O^{\prime }{A}^2=A{C}^2+O^{\prime }{C}^2$
$\Rightarrow 3^2=A{C}^2+(4-x{)}^2$
$\Rightarrow 9=A{C}^2+16+x^2-8x$
$\Rightarrow A{C}^2=-x^2-7+8x...\left(2\right)$

From equations (1) and (2), we obtain $25-x^2=-x^2-7+8x$
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.

$A{C}^2=25-x^2=25-4^2=25-16=9$
$\therefore AC=3cm$
Length of the common chord AB = 2 AC = $2\times 3$ cm = 6 cm