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

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Detailed Solution

Consider,

Proof :

Join P and Q, PQ is the chord of the circle centered at O and it is also the chord of the circle centered at O’.
Let OO' intersect PQ at M.
In
Δ OPO' and Δ OQO', we have
OP = OQ (radii of same circle)
O'P = O'Q (radii of same circle)
OO' = OO' (common)
Δ OPO' ≅ Δ OQO'(SSS congruency)
∠ POO' = ∠ QOO'
∠ POM = ∠ QOM -----(i)
Now, in Δ POM and Δ QOM we have
OP = OQ (radii of same circle)
∠ POM = ∠ QOM
OM = OM (common)
△ POM ≅ △ QOM (SAS congruency)
PM = QM and
∠ PMO = ∠ QMO
But
∠ PMO + ∠ QMO = 180 ∘ (linear pair)
2 ∠ PMO = 180
∠ PMO = 90
Hence, OO' is perpendicular bisector.