Question

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

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Answer 1

Step-by-step explanation:

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Answer 2

Question:-

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

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$\huge\mathfrak{\underline{Answer:}}$

According to the question:-

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) .

∴ ΔOAO' ≅ ΔOBO' (SSS congruency) .

∠AOO' = ∠BOO'.

⇒ ∠AOM = ∠BOM .....(1).

Now, in ΔAOM and ΔBOM, we have

⇒ OA = OB (radii of same circle) .

⇒ ∠AOM = ∠BOM 

⇒ OM = OM (common side) .

∴ ΔAOM ≅ ΔBOM (SAS congruncy) .

 AM = BM and ∠AMO = ∠BMO.

∠AMO + ∠BMO = 180° .

⇒ 2∠AMO = 180° .

⇒ ∠AMO = 90° .

Thus, AM = BM and ∠AMO = ∠BMO = 90° .

Hence, OO' is the perpendicular bisector of AB.

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