Question

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

Answer 1

Explanation:

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.