Question

if two circles intersect at two points prove that their centres 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.

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.

Hope it helps

Mark me brainlist

Answer 2

Hello mate =_=

____________________________

Solution:

Construction:

1) Draw two circles with centres O and O'.

2)Join A and B to get a common chord AB.

3) Join O and O' with the mid-point M of AB.

To prove: Centres lie on the perpendicular bisector of the common chord. In other words, we need to prove that OO' is a straight line and ∠AMO=∠AMO′=90°

In △AOB, M is the mid-point of chord AB.

⇒∠AMO=90°        .....(1)

(The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.)

Similarly, in △AO′B, M is the mid-point of chord AB.

⇒∠AMO′=90°        .......(2)

(The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.)

hope, this will help you.

Thank you______❤

_____________________________❤