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Sagot :
Answer: (0,0)
Step-by-step explanation:
[tex]\displaystyle\\\left \{ {{(x-2)^2+y^2=4} \atop {(x+2)^2+y^2=4}} \right. \\Hence,\\(x-2)^2+y^2=(x+2)^2+y^2\\(x-2)^2=(x+2)^2\\x^2-2*x*2+2^2=x^2+2*x*2+2^2\\-4x=4x\\-4x+4x=4x+4x\\0=8x\\[/tex]
Divide both parts of the equation by 8:
[tex]0=x[/tex]
Hence,
[tex](0+2)^2+y^2=4\\2^2+y^2=4\\4+y^2=4\\y^2=0\\y=0\\Thus,\ (0,0)[/tex]
Answer:
The two circles intersect at one and only point A(0 , 0)
Step-by-step explanation:
Let ς₁ be the circle of equation :
ς₁ : (x - 2)² + y² = 4
and ς₂ be the circle of equation :
ς₂ : (x + 2)² + y² = 4
Consider the point M (x , y) ∈ ς₁∩ς₂
M ∈ ς₁ ⇔ (x - 2)² + y² = 4
M ∈ ς₂ ⇔ (x + 2)² + y² = 4
M (x , y) ∈ ς₁∩ς₂ ⇒ (x - 2)² + y² = (x + 2)² + y²
⇒ (x - 2)² = (x + 2)²
⇒ x² - 4x + 4 = x²+ 4x + 4
⇒ - 4x = 4x
⇒ 8x = 0
⇒ x = 0
Substitute x by 0 in the first equation:
(0 - 2)² + y² = 4
⇔ 4 + y² = 4
⇔ y² = 0
⇔ y = 0
Conclusion:
The two circles intersect at one and only point A(0 , 0).
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