Question

If two circle intersect at two points then prove that their centres lie on the perpendicular bisector of common chord

Answer 1

HEY MATE,

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.

Hope this helps u!!!

Answer 2

Hello mate =_=

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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______❤

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