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

Q 1. Two circles of radii $5$ cm and $3$ cm intersect at two points and the distance between their centres is $4$ cm. Find the length of the common chord.

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Given: Two circles of radii $5$ cm and $3$ cm intersect at two points and the distance between their centres is $4$ cm.

To find the length of the common chord.

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

$\therefore P M \perp A B$
$\angle P M A=90^{\circ}$
Let, $\mathrm{PM}=\mathrm{x}$, so $\mathrm{QM}=4-\mathrm{x}$
In $\triangle$ APM, using Pythagoras theorem
$A M^2=A P^2-P M^2$-------(1)
Also,
In $\triangle A Q M$, using Pythagoras theorem
$A M^2=A Q^2-M Q^2$-----(2)

From 1 and 2, we get

$A P^2-P M^2=A Q^2-M Q^2$
$\Rightarrow 3^2-x^2=5^2-(4-x)^2$
$\Rightarrow 9-x^2=25-16-x^2+8 x$
$\Rightarrow 9=9+8 x$
$\Rightarrow 8 x=0$
$\Rightarrow x=0$
Put $x=0$ in equation 1

$A M^2=3^2-0^2=9$
$\Rightarrow A M=3$
$\Rightarrow A B=2 A M=6$

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