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Q : 1 Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points and the distance between their centres is $\small 4\hspace{1mm}cm$ . Find the length of the common chord.

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Given: Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points and the distance between their centres is $\small 4\hspace{1mm}cm$ .

To find the length of the common chord.

Construction: Join OP and draw $OM\perp AB\, \, and \, \, \, ON\perp CD.$

Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.

$\therefore PM\perp AB$

$\angle PMA=90 \degree$

Let, PM = x , so QM=4-x

In $\triangle$ APM, using Pythagoras theorem

$AM^2=AP^2-PM^2$ ...........................1

Also,

In $\triangle$ AQM, using Pythagoras theorem

$AM^2=AQ^2-MQ^2$ ...........................2

From 1 and 2, we get

$AP^2-PM^2=AQ^2-MQ^2$

$\Rightarrow 3^2-x^2=5^2-(4-x)^2$

$\Rightarrow 9-x^2=25-16-x^2+8x$

$\Rightarrow 9=9+8x$

$\Rightarrow 8x=0$

$\Rightarrow x=0$

Put,x=0 in equation 1

$AM^2=3^2-0^2=9$

$\Rightarrow AM=3$

$\Rightarrow AB=2AM=6$

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