Respuesta :
[tex]r=2\cos\theta[/tex]
[tex]r^2=2r\cos\theta[/tex]
[tex]\implies x^2+y^2=2x[/tex]
[tex]x^2-2x+y^2=0[/tex]
[tex]x^2-2x+1+y^2=1[/tex]
[tex](x-1)^2+y^2=1[/tex]
[tex]r^2=2r\cos\theta[/tex]
[tex]\implies x^2+y^2=2x[/tex]
[tex]x^2-2x+y^2=0[/tex]
[tex]x^2-2x+1+y^2=1[/tex]
[tex](x-1)^2+y^2=1[/tex]
Answer:
The correct option is D. (x - 1)² + y² = 1
Step-by-step explanation:
Given the equation in polar coordinates : r = 2 cosθ
To change the given equation in rectangular coordinates, we use the relation : x = r cosθ and y = r sinθ
⇒ x² + y² = r²
Now, r = 2 cosθ
Multiplying by r on both the sides
⇒ r² = 2r cosθ
⇒ x² + y² = 2x
⇒ x² - 2x + y² = 0
Making the variable x, the complete square by adding 1 on both the sides
⇒ x² - 2x + 1 + y² = 0 + 1
⇒ (x - 1)² + y² = 1
Hence, This is our required rectangular coordinate form of the given equation.
Therefore, The correct option is D. (x - 1)² + y² = 1