Question

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

Answer 1

Given : AB is the common chord of two intersecting circles (O, r) and (O′, r′). 

To Prove : Centres of both circles lie on the perpendicular bisector of chord AB, i.e., AB is bisected at right angle by OO′. 

Construction : Join AO, BO, AO′ and BO′. 

Proof : In ∆AOO′ and ∆BOO′ 

AO = OB (Radii of the circle (O, r) 

AO′ = BO′ (Radii of the circle (O′, r′)) 

OO′ = OO′ (Common) 

∴ ∆AOO′ ≅ ∆BOO′ (SSS congruency) 

⇒ ∠AOO′ = ∠BOO′ (CPCT) 

Now in ∆AOC and ∆BOC 

∠AOC = ∠BOC (∠AOO′ = ∠BOO′) 

AO = BO (Radii of the circle (O, r)) 

OC = OC (Common)

∴ ∆AOC ≅ ∆BOC (SAS congruency) 

⇒ AC = BC and ∠ACO = ∠BCO ...(i) (CPCT) 

⇒ ∠ACO + ∠BCO = 180° ..(ii) (Linear pair)