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18/07/2021 11:45 am
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If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
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18/07/2021 11:49 am
It is given that two circles intersect each other at P and Q.
To prove:
OO’ is perpendicular bisector of PQ.
Proof:
Triangle ΔPOO’ and ΔQOO’ are similar by SSS congruency since
OP = OQ and O’P = OQ (Since they are also the radii)
OO’ = OO’ (It is the common side)
So, It can be said that ΔPOO’ ΔQOO’
∴ POO’ = QOO’ — (i)
Even triangles ΔPOR and ΔQOR are similar by SAS congruency as
OP = OQ (Radii)
POR = QOR (As POO’ = QOO’)
OR = OR (Common arm)
So, ΔPOR ΔQOR
∴ PRO = QRO
Also, we know that
PRO + QRO = 180°
Hence, PRO = QRO = 180°/2 = 90°
So, OO’ is the perpendicular bisector of PQ.