Hola guys!
Question--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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Answers
Answer:
okay dear
Step-by-step explanation:
Question:−
★ If two circles intersect in two points prove that the line through their centres is the perpendicular bisector of the common chord.
\bf\underline{\underline{\blue{Given:-}}}
Given:−
★ Two circles C(O,r) and C(O' , s) intersecting at points A and B.
\bf\underline{\underline{\red{To\:Prove:-}}}
ToProve:−
★ OO', is the perpendicular bisector of AB.
\bf\underline{\underline{\green{Construction:-}}}
Construction:−
★ Draw the line segment OA, OB, O'A and O'B. Let OO' and AB intersects at M.
\bf\underline{\underline{\orange{Proof:-}}}
Proof:−
In ∆<OAO' and ∆OBO' , we have
OA = OB [each equal to r]
O'A = O'B [each equal to s]
OO' = OO' [common]
∴ ∆OAO' ≅ ∆OBO' [SSS-congruence]
==> ∠AOO' = ∠BOO'
==> ∠AOM = ∠BOM [ ∠AOO' = ∠AOM and ∠BOO' = ∠BOM]. ...(i)
In ∆AOM and ∆BOM, we have
OA = OB. [ each equal to r]
∠AOM = ∠BOM [ from (i) ]
OM = OM [comon]
∴ ∆AOM ≅ ∆BOM
==> AM = BM and ∠AMO = ∠BMO
==> AM = BM and ∠AMO = ∠BMO = 90°
==> OO' is the perpendicular bisector of AB
Answer:
hola !!
Step-by-step explanation:
ANSWER
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