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

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Given:

Two circles intersect at two points

To do:

We have to prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

Let two circles with centres $O$ and $O\'$ intersect each other at $A$ and $B$.

$OA = OB$                (Radii of the circle)

$O’A = O\'B$               (Radii of the circle)

$OO’ = OO’$              (Common side)

Therefore, by SSS congruency,

$\triangle AOO’$ and $\triangle BOO’$ are similar.

This implies,

$\triangle AOO’ \cong \triangle BOO’$

$\angle AOO’ = \angle BOO’$............(i)

In $\triangle AOC$ and \triangle BOC$,$OA = OB$(Radii)$\angle AOC = \angle BOC$($\angle AOO’ = \angle BOO’$)$OC = OC$(Common side) Therefore, by SAS congruency,$\triangle AOC \cong \triangle BOC$This implies,$\angle ACO = \angle BCO\angle ACO+\angle BCO = 180^o2\angle ACO=180^o\angle ACO = \angle BCO = \frac{180^o}{2} = 90^oOO’$is the perpendicular bisector of$AB\$.

Hence, their centres lie on the perpendicular bisector of the common chord.

Updated on 10-Oct-2022 13:46:32