Question

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

Answer 1

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

o

∴ AP=BP and ∠APO=∠BPO=90

o

∴ OO

′

is perpendicular bisector of AB

Answer 2

Answer:

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