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