Question

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

Answer 1

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

Join A, B to form the common chord \(\overline {AB}\). 

Let ‘O’ be the midpoint of AB.

Join ‘O’ with P and Q. 

Now in ΔAPO and ΔBPO 

AP = BP (radii) 

PO = PO (common) 

AO = BO (∵ O is the midpoint) 

∴ ΔAPO ≅ ΔBPO (S.S.S. congruence) 

Also ∠AOP = ∠BOP (CPCT) 

But these are linear pair of angles. 

∴ ∠AOP = ∠BOP = 90° 

Similarly in ΔAOQ and ΔBOQ 

AQ = BQ (radii) 

AO = BO (∵ O is the midpoint of AB) 

OQ = OQ (common) 

∴ AAOQ ≅ ABOQ 

Also ∠AOQ = ∠BOQ (CPCT) 

Also ∠AOQ + ∠BOQ = 180° (linear pair of angles)

∴ ∠AOQ = ∠BOQ = 180°/2 = 90°

Now ∠AOP + ∠AOQ = 180° 

∴ PQ is a line. 

Hence the proof.