Question

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

Answer 1

Answer:

ANSWER

Two circle with centre O and O′ intersect at A and B. AB is common chord of two circle OO′ is the line joining centre 

Let OO′ intersect AB at P

In OAO and OBO′ we have 

OO′→ common

OA=OB→(radii of the same circle)

O′A=O′B→(radii of the same circle)

⇒ △OAO′≅△OBO′ {SSS conguence}

∠AOO′=∠BOO′ (CPCT)

i.e., ∠AOP=∠BOP

In △AOP and BOP we have OP=OP common

∠AOP=∠BOP (proved above)

OA=OB (Radii of the semicircle)

△APD=△BPD (SSS conguence)

AP=CP (CPCT)

and ∠APO=∠BPO (CPCT)

But ∠APO+∠BPO=180

∴ ∠APO=90o

∴ AP=BP and ∠APO=∠BPO=90o

∴ OO′ is perpendicular bisector of AB