Question

If two circles intersect in two points prove that the line through their centers is the perpendicular bisector of the common chord ​

Answer 1

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Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles and OO' is the line segment joining the centres ::)

Let OO

′

intersect AB at M

Now Draw line segments OA,OB,O

′

AandO

′

B

In ΔOAO

′

andOBO

′

, we have

OA=OB (radii of same circle)

O

′

A=O

′

B (radii of same circle)

O

′

O=OO

′

(common side)

⇒ΔOAO

′

≅ΔOBO

′

(SSS congruency)

⇒∠AOO

′

=∠BOO

⇒∠AOM=∠BOM...

...(i)

Now in ΔAOM and ΔBOM we have

OA=OB (radii of same circle)

∠AOM=∠BOM (from (i))

OM=OM (common side)

⇒ΔAOM≅ΔBOM (SAS congruncy)

⇒AM=BM and∠AMO=∠BMO

But

∠AMO+∠BMO=180°

⇒2∠AMO=180°

⇒∠AMO=90°

Thus,AM=BM and∠AMO=∠BMO=90°

HenceOO

′

SO THE REQUIRED ANS IS :

is the perpendicular bisector of AB.

Answer 2

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Given: Two circles, with centres O and O' intersect at two points A and B. A AB is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect AB at P.

To prove: OO' is the perpendicular bisector of AB.

Construction: Join OA, OB, O'A and O'B

Proof: In triangles OAO' and OBO', we have

OO' = OO' (Common)

OA = OB (Radii of the same circle)

O'A = O'B (Radii of the same circle)

⇒ △OAO' ≅ △OBO' (SSS congruence criterion)

⇒ ∠AOO' = ∠BOO' (CPCT)

i.e., ∠AOP = ∠BOP

In triangles AOP and BOP, we have

OP = OP (Common)

∠AOP = ∠BOP (Proved above)

OA = OB (Radio of the same circle)

∴ △AOR ≅ △BOP (By SAS congruence criterion)

⇒ AP = CP (CPCT)

and ∠APO = ∠BPO (CPCT)

But ∠APO + ∠BPO = 180° (Linear pair)

∴ ∠APO + ∠APO = 180° ⇒ 2∠APO = 180°

⇒ ∠APO = 90°

Thus, AP = BP and ∠APO = ∠BPO = 90°

• Hence, OO' is the perpendicular bisector of AB.

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