Question

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 circle, is:
(a) root 2r
(b) rootr
c) rootЗr
d) Root 3 by 2r​

Answer 1

Answer:

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

−(

2

r

)

2

DF=

2

3

r

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

DE=

3

r

option D is the answer.

Answer 2

Answer:

c) root3r

Step-by-step 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

−(

2

r

)

2

DF=

2

3

r

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

DE= 3r