Question

Rewrite in rectangular form: r = 6 cos θ.
(1 point)
Responses

left-parenthesis x plus 3 right-parenthesis squared plus y squared equals 9
Image with alt text: left-parenthesis x plus 3 right-parenthesis squared plus y squared equals 9

left-parenthesis x minus 3 right-parenthesis squared plus y squared equals 6
Image with alt text: left-parenthesis x minus 3 right-parenthesis squared plus y squared equals 6

left-parenthesis x minus 3 right-parenthesis squared minus y squared equals 6
Image with alt text: left-parenthesis x minus 3 right-parenthesis squared minus y squared equals 6

left-parenthesis x minus 3 right-parenthesis squared plus y squared equals 9

Answer 1

To convert the polar equation \( r = 6 \cos \theta \) to rectangular form, we can use the relationships from polar to rectangular coordinates:

1. \( x = r \cos \theta \)
2. \( y = r \sin \theta \)
3. \( r^2 = x^2 + y^2 \)

From the equation \( r = 6 \cos \theta \), we can multiply both sides by \( r \):

\[
r^2 = 6r \cos \theta
\]

Substituting the polar-to-rectangular relationships into the equation:

\[
x^2 + y^2 = 6x
\]

Rearranging the equation gives us:

\[
x^2 - 6x + y^2 = 0
\]

To put it in a more recognizable form, we can complete the square for the \( x \) terms:

\[
(x^2 - 6x + 9) + y^2 = 9
\]

This simplifies to:

\[
(x - 3)^2 + y^2 = 9
\]

Thus, the rectangular form of the equation is:

\[
\boxed{(x - 3)^2 + y^2 = 9}
\]