Question

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

Answer 1

Two circle with center O and O` intersect at A and B. AB is common cord of two circles OO` is the line joining the center.

Let OO` intersect AB at P

In OAO and OBO` we have

OO` → common

OA = OB → (radious of the same circle)

O`A = O`B → (radiois of the same circle)

⇒ ∆OAO` ~ ∆OBO` {sss congruncy}

∠AOO` = ∠BOO` (CPCT)

∠AOP = ∠BOP

In ∆AOP and ∆BOP we have OP = OP (common)

∠AOP = ∠BOP (proved above)

OA = OB (radius of the semi - circle)

∆APD = ∆BPD (sss conguency)

AP = CP (CPCT)

and ∠APO = ∠BPO (CPCT)

But ∠APO + ∠BPO = 180°

∴ ∠APO = 90°

∴ AP = BP and ∠APO = ∠BPO = 90°

∴ OO` is perpendicular bisector of AB