Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm.

Solution:



Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.

$O A=O B=5 \mathrm{~cm}$

$O^{\prime} A=O^{\prime} B=3 \mathrm{~cm}$

OO' will be the perpendicular bisector of chord $A B$.

$\therefore A C=C B$

It is given that, $O O^{\prime}=4 \mathrm{~cm}$

Let $O C$ be $x$. Therefore, $O^{\prime} C$ will be $x-4$

In $\triangle O A C$

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

$\Rightarrow 5^{2}=A C^{2}+x^{2}$

$\Rightarrow 25-x^{2}=A C^{2} \ldots(1)$

In $\Delta \mathrm{O}^{\prime} \mathrm{AC}$,

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

$\Rightarrow 3^{2}=\mathrm{AC}^{2}+(x-4)^{2}$

$\Rightarrow 9=A C^{2}+x^{2}+16-8 x$

$\Rightarrow \mathrm{AC}^{2}=-x^{2}-7+8 x \ldots(2)$

From equations (1) and (2), we obtain

$25-x^{2}=-x^{2}-7+8 x$

$8 x=32$

$x=4$

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.



Length of the common chord $A B=2 O^{\prime} A=(2 \times 3) \mathrm{cm}=6 \mathrm{~cm}$