Question

convert the polar equation $r = 4cos\theta$ into a rectangular equation. example solution input should an equation in terms of $x$ and $y$. if the solution is $y = x^2 + x$ then input $y=x^2+x$. a solution write - up is required for this problem. answer:

Explanation:

Step1: Recall polar-rectangular relations

We know that \( x = r\cos\theta \) and \( r^{2}=x^{2}+y^{2} \). Given the polar equation \( r = 4\cos\theta \).

Step2: Multiply both sides by \( r \)

Multiply each side of \( r = 4\cos\theta \) by \( r \) to get \( r^{2}=4r\cos\theta \).

Step3: Substitute \( r^{2} \) and \( r\cos\theta \)

Substitute \( r^{2}=x^{2}+y^{2} \) and \( r\cos\theta = x \) into the equation \( r^{2}=4r\cos\theta \). We obtain \( x^{2}+y^{2}=4x \).

Step4: Rearrange the equation

Rearrange the equation \( x^{2}+y^{2}=4x \) to the standard form of a circle. Subtract \( 4x \) from both sides: \( x^{2}- 4x+y^{2}=0 \). Then complete the square for the \( x \)-terms. \( x^{2}-4x=(x - 2)^{2}-4 \). So the equation becomes \( (x - 2)^{2}-4+y^{2}=0 \), which simplifies to \( (x - 2)^{2}+y^{2}=4 \). But if we want it in terms of \( x \) and \( y \) without completing the square, we can also leave it as \( x^{2}+y^{2}-4x = 0 \) or rearrange for \( y^{2}=4x - x^{2} \), but the most direct conversion from \( r = 4\cos\theta \) using the substitutions gives \( x^{2}+y^{2}=4x \).

Answer:

\( x^{2}+y^{2}=4x \) (or the standard circle form \( (x - 2)^{2}+y^{2}=4 \))