Question

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

Answer 1

Step-by-step explanation:

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.