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

Solution:

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.

☛ Check: NCERT Solutions Class 9 Maths Chapter 10

Video Solution:

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

Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.3 Question 3

Summary:

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

☛ Related Questions:

• Prove that if chords of congruent circles subtend equal angles at their centers, then the chords are equal.
• Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
• Suppose you are given a circle. Give a construction to find its center.