Question

if two circle intersect at two points prove their Centre lie on the perpendicular bisector of the common chord​

Answer 1

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'A and O'B In ΔOAO' and OBO' , 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° Hence OO' is the perpendicular bisector of AB.

❤❤ Mark as Brainliest ❤❤

Answer 2

Answer:

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.