Question

If two circles intersect at two plonte, prove that there centres lie on the perpendicular bisectorof the common chord.​

Answer 1

Answer:

Two circles, with centres O and O' intersect at two points A and B. A AB is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect AB at P.

To prove: OO' is the perpendicular bisector of AB.

Construction: Join OA, OB, O'A and O'B

Proof: In triangles OAO' and OBO', we have

OO' = OO' (Common)

OA = OB (Radii of the same circle)

O'A = O'B (Radii of the same circle)

⇒ △OAO' ≅ △OBO' (SSS congruence criterion)

⇒ ∠AOO' = ∠BOO' (CPCT)

i.e., ∠AOP = ∠BOP

In triangles AOP and BOP, we have

OP = OP (Common)

∠AOP = ∠BOP (Proved above)

OA = OB (Radio of the same circle)

∴ △AOR ≅ △BOP (By SAS congruence criterion)

⇒ AP = CP (CPCT)

and ∠APO = ∠BPO (CPCT)

But ∠APO + ∠BPO = 180° (Linear pair)

∴ ∠APO + ∠APO = 180° ⇒ 2∠APO = 180°

⇒ ∠APO = 90°

Thus, AP = BP and ∠APO = ∠BPO = 90°

Hence, OO' is the perpendicular bisector of AB.

follow me....❣️❣️

Answer 2

hyy I need good friend can you my friend if you want;)

Maths

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

share

Share

Answer

Two circle with centre O and O

′

intersect at A and B. AB is common chord of two circle OO

′

is the line joining centre

Let OO

′

intersect AB at P

In OAO and OBO

′

we have

OO

′

→ common

OA=OB→(radii of the same circle)

O

′

A=O

′

B→(radii of the same circle)

⇒ △OAO

′

≅△OBO

′

{SSS conguence}

∠AOO

′

=∠BOO

′

(CPCT)

i.e., ∠AOP=∠BOP

In △AOP and BOP we have OP=OP common

∠AOP=∠BOP (proved above)

OA=OB (Radii of the semicircle)

△APD=△BPD (SSS conguence)

AP=CP (CPCT)

and ∠APO=∠BPO (CPCT)

But ∠APO+∠BPO=180

∴ ∠APO=90

o

∴ AP=BP and ∠APO=∠BPO=90

o

∴ OO

′

is perpendicular bisector of AB