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

Given : Circle C (P,r) and circle C (Q, r’) intersects each other at the points A and B.
To prove: points p and Q lies on the perpendicular bisector of common chord AB.
Construction: Join point P and Q to mid-point M of chord AB.
Proof: AB is chord of circle C (P, r) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.]
Hence, ∠PMA = 90°
Similarly, AB is chord of circle C (Q, r) and QM is bisector of chord AB.
Therefore, QM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.]
Hence, ∠QMA = 90
Now, ∠PMA + ∠QMA = 90° + 90° = 180°
Since, ∠PMA and ∠QMA are forming liner pair. So PMQ is a straight line.
Hence, points P and Q lies on the perpendicular bisector of common chord AB.