Question

Two congruent circles of radius r intersect such that each passes through the center 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

Hint- In order to find the solution first we will make the diagram according to the problem statement and then we will further proceed by finding some right angled triangle in order to find the relation between the length of radius and the common chord using Pythagoras theorem.

Complete step-by-step answer:

Let the two circles have their centers at A and C.

Let them intersect at D and E.

Now AC, AD, AE, CD, CA, CE all is radii of those two circles so their lengths must be r

By symmetry F is the midpoint of AC

So AF, AC must be $\dfrac{r}{2}$

Also DE and AC must intersect at right angles.

 So by Pythagoras theorem:

$A{F^2} + D{F^2} = A{D^2} $

$  D{F^2} = A{D^2} - A{F^2} $

$  D{F^2} = {r^2} - {\left( {\dfrac{r}{2}} \right)^2} $

 $ DF = \dfrac{{\sqrt 3 r}}{2} $

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

\[DE = \sqrt 3 r\]

Hence, option “D” is the answer.

Note- In order to solve these types of questions, visualization is important. In solving problems related to coordinate geometry it is necessary to draw figures. In order to solve such types of problems mostly we use the theorems giving the relation between the dimension of chord and tangent and the radius. Also it is important to find the angle between them sometime.