Question

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

Answer 1

Let the two circles with centre P and Q intersect at points A and B.

Join AB. AB is the common chord.

Join PQ. AB and PQ bisect each other at M.

Let M be the midpoint of AB.

Hence, PM ⊥ AB [Since, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord]

⇒ ∠PMA = 90º

Now, since M is the midpoint of AB

Hence, QM ⊥ AB

⇒ ∠QMA = 90º

Thus, ∠PMQ = ∠PMA + ∠QMA = 90º + 90º = 180º

Hence PMQ is a straight line and PMQ ⊥ AB

Therefore, PMQ is the perpendicular bisector of the common chord AB and passes through the two centers P and Q.

So, the centres lie on the perpendicular bisector of the common chords.

Answer 2

if two circles intersect at two points then it is obvious that their centre lie on the perpendicular bisector of the common chord

hence proved