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Answer 1

Answer:

If two circles intersect at two points, prove that their centre lie on the perpendicular bisector of the common chord.

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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