Math, asked by sdiya0804, 10 hours ago

if two circles intersect at two points prove that their centres lie on the perpendicular bisector of the common chord​

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Answered by supermicrons
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 ′

→ commonOA=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 o

∴ AP=BP and ∠APO=∠BPO=90 o

∴ OO ′ is perpendicular bisector of

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