Question

if two circles intersect at two points, prove that their centers lie on the perpendicular bisectes of common chord​

Answer 1

Answer:

ok...................

Answer 2

Hello

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 it helps you

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