Question

If two circles intersects at 2 points, prove that their centres lie oa the perpendicular
bisectors of the common chord,​

Answer 1

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=90°

∴ AP=BP and ∠APO=∠BPO=90°

∴ OO' is perpendicular bisector of AB

hope will be helpful ☺️