Question

Two equal circles of radius r intersect such that each passes through the centre of the other the length of common chord is how much

Answer 1

▶Let the two circles have their centres at AAand CC.

▶Let them intersect at DD and EE.

▶Now AC,AD,AE,CD,CA,CE all are radii of those two circles so these must be r/2

By symmetry F is the midpoint of AC so
$AF, \: AC \: must \: be \: \frac{r}{2}$

▶Also DE and AC must intersect at right angles, so by Pythagoras theorem we have :

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

▶Solving we get:
$= > DF \: \: = \frac{ \sqrt{3 \: r} }{2}$

▶So, DE must be double of DF

$DE= \sqrt{3} \: r$


▶HOPE IT HELPS YOU

Answer 2

Answer:

Step-by-step explanation:

Hope it'll help anyone