The rectangular form of [tex]\mathbf{r = 4cos(\theta)}[/tex] is (a) [tex]\mathbf{(x - 2)^2 + y^2= 4}[/tex]
The polar coordinate is given as:
[tex]\mathbf{r = 4cos(\theta)}[/tex]
Multiply both sides of [tex]\mathbf{r = 4cos(\theta)}[/tex] by r
[tex]\mathbf{r \times r= 4cos(\theta) \times r}[/tex]
Evaluate
[tex]\mathbf{r^2= 4rcos(\theta) }[/tex]
As a general equation, we have:
[tex]\mathbf{x^2 + y^2 = r^2}[/tex]
Substitute [tex]\mathbf{x^2 + y^2 = r^2}[/tex] in [tex]\mathbf{r^2= 4rcos(\theta) }[/tex]
[tex]\mathbf{x^2 + y^2= 4rcos(\theta) }[/tex]
Also as a general equation, we have:
[tex]\mathbf{x = rcos(\theta)}[/tex]
So, [tex]\mathbf{x^2 + y^2= 4rcos(\theta) }[/tex] becomes
[tex]\mathbf{x^2 + y^2= 4x}[/tex]
Subtract 4x from both sides
[tex]\mathbf{x^2-4x + y^2= 0}[/tex]
Complete the square on x^2 - 4x
[tex]\mathbf{(x - 2)^2-2 + y^2= 0 + 2}[/tex]
Add 2 to both sides
[tex]\mathbf{(x - 2)^2 + y^2= 0 + 2+2}[/tex]
[tex]\mathbf{(x - 2)^2 + y^2= 4}[/tex]
Hence, the rectangular form of [tex]\mathbf{r = 4cos(\theta)}[/tex] is (a) [tex]\mathbf{(x - 2)^2 + y^2= 4}[/tex]
Read more about rectangular forms at:
https://brainly.com/question/11803861
