Question

If two circles intersect at two points , prove that their centers lie on the perpendicular bisector 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

Explanation:

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