Question

if two circle intersect at two points prove that their Centre lies on the perpendicular bisector of the common chord

Answer 1

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Answer 2

Join the common chord. Then let mark the chord AB and the center of two circles be X and Y , respectively.L et the point be named O where it meets at chord AB.


In triangle AXY and BXY,
AX=BX (Radius of the same circle)
AY=BY (Radius of the same circle)
XY=XY. (Common side)
By SSS, ∆ AXY congruent to ∆BXY.
By CPCT, angle (AXY) = angle (BXY)
AX=BX

In ∆AOX And ∆BOX,
AX=BX. (By CPCT)
angle AXO= angle BXO ( By CPCT)
OX=OX (Common side)

By SAS, ∆AOX Congruent to ∆BOX
By CPCT, angle AOX =angle BOX
OA=OB

So, angle AOX + angle BOX =180°
angle AOX + angle AOX=180°
2(angle AOX)=180°
angle AOX=90°

So, their centres lie on perpendicular bisector of common chord.