Question

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

Answer 1

Answer:

please refer to the attachment.. hope it helps you

Answer 2

Answer:

Step-by-step explanation:

Consider two circles centered at point O and O’, intersecting each other at point A and B respectively.

Join A and B, AB is the chord of the circle centered at O and it is also the chord of the circle centered at O’.

Let OO' intersect AB at M.

In ΔOAO' and ΔOBO', we have

OA = OB (radii of same circle)

O'A = O'B (radii of same circle)

OO' = OO' (common)

ΔOAO' ≅ ΔOBO'(SSS congruency)

∠AOO' = ∠BOO'

∠AOM = ∠BOM -----(i)

Now, in ΔAOM and ΔBOM we have

OA = OB (radii of same circle)

∠AOM = ∠BOM

OM = OM (common)

△AOM ≅ △BOM (SAS  congruency)

AM = BM and ∠AMO = ∠BMO

But

∠AMO + ∠BMO = 180∘ (linear pair)

2∠AMO =  180∘

∠AMO = 90∘

Hence, OO' is perpendicular bisector.