Question

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

Answer 1

To prove: $CD\perp AB$ and $AO=OB$

Given: $AC=BC=r_1$

$AD=BD=r^2$

Proof: Consider circle $C_1$, $AB$ is chord of $C_1$

perpendicular from center to chord $AB$ bisects the chord

$\Rightarrow$ $CD\perp AB$ and $AO=OB$

$\Rightarrow$ $\angle COA={90}^o$

Conisider circle $C_2$

$AB$ is chord of $C_2$

$\Rightarrow$ $DO\perp AB$ and $AO=OB$

$\angle DOA={90}^o.......\left(2\right)$

From (1) and (2) $\angle DOA+\angle COA=\angle DOC={180}^o$

$\therefore$ $CD$ is a straight line

perpendicular to $AB$

$\Rightarrow$ $CD\perp AB$ and $AO=OB$