Question

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

Answer 1

Given :-

2 circles

circle with radius r and center o and

circle with radius r` and center o` ( ` This is just a dash to differentiate between 2 circles )

AB is common cord

To prove :-

OO` is perpendicular bisector

Construction :-

Join OA, OB , O`A ,O`B

Proff:-

In triangles OAO` and OBO`

OA=OB ( radii of same circle )

O`A =O`B ( radii of same circle)

OO` = OO` ( Common )

Therefore by SSS Congrency

Triangle OPO` is congrent to Triangle OBO`

Angle 1 = Angle 2 ( C.P.C.T)

NOW,

In triangles OPA and OMB

OA = OB ( radii of same circle )

Angle 1 = Angle 2 ( Proved above )

OM = OM ( Common )

Therefore by SAS Congrency

Triangle OMA is congrent to Triangle OMB

Therefore AM = MB ( C.P.C.T)

So OO` bisects AB

Also Angle 3 = Angle 4 ( C.P.C.T)

NOW,

Angle 3 + Angle 4 =180 ( LINEAR PAIR )

Angle 3 + Angle 3 = 180 ( As Angle 3 = Angle 4)

2 Angle 3 = 180

Angle 3 = 180/2

Angle 3 = 90

Therefore,

Angle 3 = Angle 4 = 90

So OO` IS PERPENDICULAR ON AB

THEREFORE OO` IS PERPENDICULAR BISECTOR OF AB

Note :-

Figure in attachment

Answer 2

Step-by-step explanation:

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