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Do the circles having the equations (x - 2)2 + y2 = 4 and (x + 2)2 + y2 = 4 intersect on the Cartesian plane? If so, where do they intersect?

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]

View image Sangers1959

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).

View image Profarouk
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