Question

if two circles intersect at two points then prove that the circles lie on the perpendicular bisector of the common chord
can any one plz answer this plz with pic I'll mark u brainiest. plzzzzzz

Answer 1

HERE IS YOURS SOLUTION;

◆ 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 ★

HENCE PROVED

HOPE IT HELPS