Let's convert the given polar equation [tex]\( r = 4 \cos(\theta) \)[/tex] to its rectangular form.
Recall the relationships between polar coordinates [tex]\((r, \theta)\)[/tex] and rectangular coordinates [tex]\((x, y)\)[/tex]:
1. [tex]\( x = r \cos(\theta) \)[/tex]
2. [tex]\( y = r \sin(\theta) \)[/tex]
3. [tex]\( r = \sqrt{x^2 + y^2} \)[/tex]
4. [tex]\(\cos(\theta) = \frac{x}{r}\)[/tex]
Now, let's substitute these relationships into the given polar equation:
[tex]\[ r = 4 \cos(\theta) \][/tex]
We can express [tex]\(\cos(\theta)\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(r\)[/tex]:
[tex]\[ r = 4 \left(\frac{x}{r}\right) \][/tex]
Multiply both sides by [tex]\( r \)[/tex] to eliminate the fraction:
[tex]\[ r^2 = 4x \][/tex]
We know that [tex]\( r^2 = x^2 + y^2 \)[/tex]. Therefore:
[tex]\[ x^2 + y^2 = 4x \][/tex]
Now, let's rearrange this equation to match the standard form of a circle equation. To do so, we complete the square for the [tex]\( x \)[/tex] terms.
First, rewrite the equation:
[tex]\[ x^2 + y^2 = 4x \][/tex]
Subtract [tex]\( 4x \)[/tex] from both sides:
[tex]\[ x^2 - 4x + y^2 = 0 \][/tex]
To complete the square for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 4x + y^2 = 0 \][/tex]
Add and subtract [tex]\( 4 \)[/tex] inside the equation:
[tex]\[ x^2 - 4x + 4 + y^2 = 4 \][/tex]
This can be written as:
[tex]\[ (x - 2)^2 + y^2 = 4 \][/tex]
Thus, the rectangular form of the given polar equation [tex]\( r = 4 \cos(\theta) \)[/tex] is:
[tex]\[ (x-2)^2 + y^2 = 4 \][/tex]
Therefore, the correct option is:
[tex]\[ (x-2)^2+y^2=4 \][/tex]
