Question

If two circles interested at two points, then prove that their centers lie on the perpendicular bisector of common chord. ​

Answer 1

Answer:

Step-by-step explanation:

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, Join the 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) .

$\therefore$ ΔOAO' ≅ ΔOBO' (SSS congruency) .

$\because$ ∠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) .

$\therefore$ ΔAOM ≅ ΔBOM (SAS congruncy) .

$\because$ AM = BM and ∠AMO = ∠BMO.

But,

$\because$ ∠AMO + ∠BMO = 180° .

⇒ 2∠AMO = 180° .

⇒ ∠AMO = 90° .

Thus, AM = BM and ∠AMO = ∠BMO = 90° .

Hence, OO' is the perpendicular bisector of AB.

Answer 2

Solution:-

To prove :-

→centers lie on the perpendicular bisector of common chord.

$\angle \: AMO \: = \angle \: AMO' = 90 \degree \:$

Construction:-

1. Draw two circles with centre O and O' respectively .
2. Join the points A and B to get a common chord (AB)
3. Join center's of circle at point M on common chord AB

Now,

In ∆ AOB ,M is the mid point of AB

$\implies \: \angle \: AMO=90 \degree \: \: .........(1) \\ \\ \because \: \: M \: is \: perpendicular \: on \: O \: and \: O'$

Similarly in ∆ AO'B. , M is the mid point of AB .

$\therefore \: \angle \: AMO'= 90 \degree$

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

Hence proved.