Two equal circles of radius `r` intersect such that each passes through the centre of the other. The length of the common chord of the circles is `sqrt(r)` (b) `sqrt(2)\ r\ A B` (c) `sqrt(3)\ r\ ` (d) `(sqrt(3))/2\ r`Let the two circles have their centres at `A and C.`

Explanation

Let the two circles have their centres at `A and C.`
Let them intersect at `D and E.`
Now `AC,AD,AE,CD,CA,CE` all are radii of those two circles so their lengths must be `r`
By symmetry `F` is the midpoint of `AC`.
So `AF,AC` must be `r/2`
Also `DE \and\ AC` must intersect at right angles.
So by Pythagoras theorem:`AF^2+DF^2=AD^2`
`DF^2=AD^2−AF^2`
`DF^2=r^2−(r/2)C^2`
`DF=sqrt3/2r`
So,`DE` must be double of `DF`(Again, by symmetry

`DE=(sqrt3)r`