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°

∴ AP=BP and ∠APO=∠BPO=90°

∴ OO ′ is perpendicular bisector of AB

Answer 2

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′ → commonOO′→common

OA=OB→OA=OB→ (radii of the same circle)

O′ A=O′ B→O′A=O′B→ (radii of the same circle)

⇒ △OAO′ ≅△OBO′⇒△OAO′≅△OBO′ {SSS conguence}

∠AOO′ =∠BOO′ (CPCT)∠AOO′=∠BOO′(CPCT)

i.e., ∠AOP=∠BOPi.e.,∠AOP=∠BOP

In △AOP and BOP we have OP=OP common

∠AOP=∠BOP∠AOP=∠BOP (proved above)

OA=OBOA=OB (Radii of the semicircle)

△APD=△BPD△APD=△BPD (SSS conguence)

AP=CP (CPCT)AP=CP(CPCT)

and ∠APO=∠BPO (CPCT)∠APO=∠BPO(CPCT)

But∠APO+∠BPO=180∠APO+∠BPO=180

∴ ∠APO=90°

∴ AP=BP and ∠APO=∠BPO=90°

∴ OO ′ is perpendicular bisector of AB