Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.


Given :

Radii of two circles are 5 cm and 3 cm.

The distance between  the center of the circles is 4 cm.

To do :

We have to find the length of the common chord.

Solution : 


                   


In the above figure, $AO=5 cm, BO=3 cm$

$AB = 4 cm, AC = x, BC = 4-x$.

$OD$ is the common chord of two circles.

We have to find the length of common chord $OD$.

We know that,

The perpendicular bisector of the chord passes through the center of the circle.

So, $OC = CD$ and $\angle ACO =\angle BCO = 90°$.

In $\triangle ACO, AO^2 = AC^2+CO^2$

                               $5^2 =x^2+CO^2$

                               $CO^2 = 5^2-x^2$

                               $CO^2=25-x^2$.............(i)

In $\triangle BCO, BO^2 = BC^2+CO^2$

                                $3^2=(4-x)^2+CO^2$

                                $CO^2=3^2-(4-x)^2$

                                $CO^2 = 9-16+8x-x^2$.........(ii)

Equate (i) and (ii), we get,

$25-x^2= 9-16+8x-x^2$

$25 = -7+8x$

$8x = 25+7$

$8x=32$

$x = \frac{32}{8}=4$

Substitute $x=4$ in (i),

 $CO^2=25-4^2$

 $CO^2=25-16$

$CO^2=9$

$CO=3$

We know that, $CO=CD=3$.

$OD=OC+CD=3+3=6$

$OD=6 cm$.

Therefore, the length of the common chord is 6 cm. 



Tutorialspoint
Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

135 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements