Question

If two circles intersect at any two points prove that their centres lie on thr perpendicular

Answer 1

Correct Question :

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

Answer :

Given -

• AB is the common chord intersecting circles at A and B with centre O and O'.

Proof -

$\\ \sf \: AO = AO' \\ \qquad \quad \qquad \sf \bigg (radii \: \: of \: \: the \: \: same \: \: circle \bigg) \\ \sf BO = BO' \\ \\ \sf \: OO' = OO' \qquad \quad \: (common) \\ \\ \\ \therefore \: \: \triangle \sf \: AOO' \cong \: \triangle \: \sf \: BOO' \qquad \: (sss) \\ \\ \\ \sf \angle \: 3 = \: \angle \: 4 \qquad \: ( \: by \: \: cpct \: ) \: - - - (1)$

Now, in ∆ AOP and ∆ BOP

$\\ \sf \: AO = OB \qquad \: (radii \: \: of \: \: the \: \: same \: \: circle \: ) \\ \\ \sf \: OP = OP \qquad \: ( \: common \: ) \\ \\ \sf \: \angle \: 3 = \: \angle \: 4 \qquad \: ( \: from \: (1) \: )$

_______________________________________

$\\ \sf \triangle \: \: AOP \: \cong \: \triangle \: BOP \qquad \: (sas) \\ \\ \therefore \sf \: AP = PB \\ \\ \sf \angle \: 1 = \angle \: 2 \qquad \: (cpct)$

Also,

$\\ \sf \angle \: 1 + \angle \: 2 = 180 {}^{ \circ} \\ \\ \\ \implies \sf \: 2 \: \angle \: 1 = 180 {}^{ \circ} \\ \\ \\ \implies \sf \: \angle \: 1 = 90 {}^{ \circ}$