Question

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?

Answer 1

Answer: (0,0)

Explanation:

`\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\\`

Divide both parts of the equation by 8:

`0=x`

Hence,

`(0+2)^2+y^2=4\\2^2+y^2=4\\4+y^2=4\\y^2=0\\y=0\\Thus,\ (0,0)`

Answer 2

The two circles intersect at one and only point A(0 , 0)

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).