If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common - Maths - Circles

Question

If two circles intersect at two points, then prove that
their centres lie on the perpendicular bisector of the common
chord

Koka Sri Lakshmi Divya Sai · Accepted Answer

${Let two circles with centers O and O}^{1} intersect at two points A and B {so that AB is the common chord of two circles and OO}^{1} is the line segment joining the centres of the two circles {Let OO}^{1} intersect AB at M {Now Draw line segments OA ,OB,O}^{1} {A and O}^{1} B In Δ {OAO}^{1} {and OBO}^{1}, we have OA=OB (radii of same circle) O^{1} A= O^{1} B (radii of same circle) O^{1} O= O O^{1} (common side) \Rightarrow Δ {OAO}^{1} ≅ Δ {OBO}^{1} (SSS congruency) \Rightarrow ∠ {AOO}^{1} = ∠ {BOO}^{1} \Rightarrow ∠ AOM= ∠ BOM - - - - (i) Now in Δ AOM and BOM we have OA=OB (radii of same sircle) ∠ AOM= ∠ BOM (from (i)) OM=OM (commom side) \Rightarrow Δ AOM ≅ Δ BOM (SAS congruncy) \Rightarrow AM=BM and ∠ AMO= ∠ BMO But ∠ AMO+ ∠ BMO = 180^{0} \Rightarrow 2 ∠ {AMO=180}^{0} \Rightarrow ∠ {AMO=90}^{0} Thus, AM =BM and ∠ {AMO= ∠ BMO=90}^{0} {Hence OO}^{1} is the perpendicular bisector of AB$

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If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.