Question

which equation represents the rectangular form of r = 4cos(θ)? (x - 2)^2 + y^2 = 4 (x + 2)^2 + y^2 = 4 (x - 2)^2 + y^2 = 0 (x + 2)^2 + y^2 = 0

Explanation:

Step1: Recall the conversion formulas

$x = r\cos\theta$, $y = r\sin\theta$, and $r^{2}=x^{2}+y^{2}$

Step2: Multiply both sides by r

Given $r = 4\cos\theta$, we get $r^{2}=4r\cos\theta$

Step3: Substitute conversion formulas

Since $r^{2}=x^{2}+y^{2}$ and $x = r\cos\theta$, we have $x^{2}+y^{2}=4x$

Step4: Rearrange the equation

$x^{2}-4x + y^{2}=0$

Step5: Complete the square for x - terms

$(x - 2)^{2}-4+y^{2}=0$, which simplifies to $(x - 2)^{2}+y^{2}=4$

Answer:

A. $(x - 2)^2 + y^2=4$