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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