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

S seema garhwal

Given: Two circles intersect at two points.

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

Construction: Joinpoint 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.

$\therefore PM\perp AB$

Hence, $\angle PMA =90 \degree$

Similarly, AB is a chord of circle(Q,r' ) and QM is the bisector of chord AB.

$\therefore QM\perp AB$

Hence, $\angle QMA =90 \degree$

Now, $\angle QMA +\angle PMA=90 \degree+90 \degree= 180 \degree$

$\angle$PMA and $\angle$QMA are forming 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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