Two circles having radii $$r$$ units intersect each other in such a way that each of them passes through the centre of the other. Then the length of their common chord is

According to question

Let the radius of the circle be = $$r$$
∴ DO = OC = $$\frac{r}{2}$$
In right angle ΔAOD
By using Pythagoras theorem
$$\eqalign{ & A{D^2} = O{D^2} + A{O^2} \cr & {r^2} = \frac{{{r^2}}}{4} + A{O^2} \cr & A{O^2} = {r^2} - \frac{{{r^2}}}{4} \cr & A{O^2} = \frac{{3{r^2}}}{4} \cr & AO = \frac{{\sqrt 3 r}}{2}\,\,\,\left( {\because AB = 2 \times AO} \right) \cr & AB = \frac{{\sqrt 3 r}}{2} \times 2 \cr & AB = \sqrt 3 r{\text{ units}} \cr} $$