If two circles intersect in two points prove that the line through their centres is the perpendicular bisector of the common chord.
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Answer:
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Explanation:
Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles and OO' is the line segment joining the centres
Let OO
′
intersect AB at M
Now Draw line segments OA,OB,O
′
AandO
′
B
In ΔOAO
′
andOBO
′
, we have
OA=OB (radii of same circle)
O
′
A=O
′
B (radii of same circle)
O
′
O=OO
′
(common side)
⇒ΔOAO
′
≅ΔOBO
′
(SSS congruency)
⇒∠AOO
′
=∠BOO
′
⇒∠AOM=∠BOM......(i)
Now in ΔAOM and ΔBOM we have
OA=OB (radii of same circle)
∠AOM=∠BOM (from (i))
OM=OM (common side)
⇒ΔAOM≅ΔBOM (SAS congruncy)
⇒AM=BM and∠AMO=∠BMO
But
∠AMO+∠BMO=180°
⇒2∠AMO=180°
⇒∠AMO=90°
Thus,AM=BM and∠AMO=∠BMO=90°
HenceOO
′
is the perpendicular bisector of AB.
MARK AS BRILLANT
Question:-
- If two circles intersect in two points prove that the line through their centres is the perpendicular bisector of the common chord.
Sólution:-
- Given:
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.
- Hope you are helped✔️
