Question

Two congruent circles of radius r intersect such that each passes through the centre of the other, then the length of the common chord is given by ?

• A. $r$
• B. $2r$
• C. $\sqrt{2}r$
• D. $\sqrt{3}r$

Answer 1

The correct option is D $\sqrt{3}r$

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:

${AF}^2+{DF}^2={AD}^2$

${DF}^2={AD}^2-{AF}^2$

${DF}^2=r^2-{\left(\frac{r}{2}\right)}^2$

$DF=\frac{\sqrt{3}}{2}r$

So, DE must be double of DF (Again, by symmetry)

$DE=\sqrt{3}r$

option D is the answer.