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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.