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if tow circle intersect at tow points, prove that their centers lie on the perpendicular bi sector of the common chord
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Proof that the centers lie on the perpendicular bisector of the common chord:




Given: Two circles intersect at two points

To Prove: The centers of the circles lie on the perpendicular bisector of the common chord




Proof:

- Let the two circles intersect at points A and B, creating a common chord AB
- Let O1 and O2 be the centers of the circles
- Connect O1A, O1B, O2A, and O2B
- As O1A = O1B and O2A = O2B, OA and OB are the radii of the circles
- This creates two congruent triangles O1AO2 and O1BO2 by SSS congruence
- Therefore, angle O1AO2 = angle O1BO2
- This means that the line joining the centers O1 and O2 is perpendicular to the common chord AB
- Hence, the centers of the circles lie on the perpendicular bisector of the common chord


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if tow circle intersect at tow points, prove that their centers lie on the perpendicular bi sector of the common chord Related: What is a Circle??
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