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  • If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Q : 3 If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
 

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Given: Two circles intersect at two points.

To prove: their centres lie on the perpendicular bisector of the common chord.

Construction: Join point P and Q to midpoint M of chord AB.

Proof: AB is a chord of circle C(Q,r)  and QM is the bisector of chord AB.

$\therefore P M \perp A B$
Hence, $\angle P M A=90^{\circ}$
Similarly, $A B$ is a chord of the circle ( $Q, r^{\prime}$ ) and $Q M$ is the bisector of chord $A B$.
$\therefore Q M \perp A B$
Hence, $\angle Q M A=90^{\circ}$
Now, $\angle Q M A+\angle P M A=90^{\circ}+90^{\circ}=180^{\circ}$
$\angle \mathrm{PMA}$ and $\angle \mathrm{QMA}$ form linear pairs so PMQ is a straight line.

Hence, P and Q lie on the perpendicular bisector of common chord AB.

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