Question

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

Answer 1

Consider two circles centered at point O and O’, intersecting each other at point A and B respectively.

Join A and B, AB is the chord of the circle centered at O and it is also the chord of the circle centered at O’.
Let OO' intersect AB at M.
In $\Delta$OAO' and $\Delta$OBO', we have
OA = OB (radii of same circle)
O'A = O'B (radii of same circle)
OO' = OO' (common)
$\Delta$OAO' $\cong$ $\Delta$OBO'(SSS congruency)
$\angle$AOO' = $\angle$BOO'
$\angle$AOM = $\angle$BOM -----(i)

Now, in $\Delta$AOM and $\Delta$BOM we have
OA = OB (radii of same circle)
$\angle$AOM = $\angle$BOM
OM = OM (common)
$△$AOM $\cong$ $△$BOM (SAS congruency)
AM = BM and $\angle$AMO = $\angle$BMO
But
$\angle$AMO + $\angle$BMO = ${180}^{\circ }$ (linear pair)
2$\angle$AMO = ${180}^{\circ }$
$\angle$AMO = ${90}^{\circ }$
Hence, OO' is perpendicular bisector.