Question

Two circles intersect each other at 2 points,prove that their centres lie on the perpendicular bisector of the common chord

Answer 1

GIVEN : Two circles with their centres at O and M intersect each other at the points P and Q respectively. Thus, PQ is their common radius.

TO PROVE : The two centres, O and M lie on the perpendicular bisector of the common chord PQ.

CONSTRUCTION : From O, draw a perpendicular on the chord PQ to the point X and extend it upto the point M which is the centre of another circle. Join O,P ; P,M ; Q,M ; O,Q.

PROOF :

In ∆ POX and ∆ OQX

OP = OQ (Radii of same circle)

<PXO = <OXQ (Each 90° since OX⊥PQ)

OX is the common side.

So, ∆ POX ≅ ∆ OQX (R-H-S rule of congruency)

Hence, PX = QX [C. P. C. T.]

We can say, OX is the perpendicular bisector of chord PQ from "O", i.e., O lies on the perpendicular bisector of PQ.

$AGAIN,$ In ∆ PXM and ∆ OXM

PX = QX (We got just before)

MP = MQ (Radii of same circle)

OX is the common side.

So, ∆ PXM ≅ ∆ QXM (S-S-S rule of congruency)

Hence, <PXM = <QXM (C. P. C. T)

Since both <PXM and <QXM lies on the common chord PQ which is a straight line, <PXM = <QXM = 90°.

We can say, MX is the perpendicular bisector of chord PQ from "M", i.e., M also lies on the perpendicular bisector of PQ.

Thus, the theorem is proved.

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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