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# Two circles of radii $ 5 \mathrm{~cm} $ and $ 3 \mathrm{~cm} $ intersect at two points and the distance between their centres is $ 4 \mathrm{~cm} $. Find the length of the common chord.

Given:

Radii of two circles are \( 5 \mathrm{~cm} \) and \( 3 \mathrm{~cm} \).

The distance between the centre of the circles is \( 4 \mathrm{~cm} \).

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 the common chord $OD$.

We know that,

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

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

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 $6\ cm$.

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