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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: 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

\anglePMA and \angleQMA 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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