Question

if 2 circles intersect at 2 points then prove that their centres lie on the perpendicular bisector of the common chord

Answer 1

To prove :OO' is the perpendicular bisector of AB
Construction :join OA, Ob, O'A and O'B
Pf:OO'=OBO'(common)
Oa=ob
OAO'=OBA'
AOO'=BOO
Aop=Bop
Op =op
Aop=BOP

OA=OB
AOR =CONGRUENT BOP
AP=BP
APO= BPO

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______❤

_____________________________❤