Converting r=4+4cos(theta) into rectangular form

[Raerin]

So how do I convert r=4+4cos(theta) into rectangular form?
I know that r^2 = x^2+y^2 and that rcos(theta) = x.

Would the start of the solution be:

sqrt(x^2 + y^2) = 4+4x

If yes, I don't know where to go from there.

 

[I like Serena]

So how do I convert r=4+4cos(theta) into rectangular form?
I know that r^2 = x^2+y^2 and that rcos(theta) = x.

Would the start of the solution be:

sqrt(x^2 + y^2) = 4 +4x

If yes, I don't know where to go from there.

Almost. You have made a small mistake.

We can write $r\cos(\theta) = x$ as
$$\cos(\theta) = \frac x r \qquad\qquad (1)$$

So you should have
$$\sqrt{x^2 + y^2} = 4 + 4 \frac{x}{\sqrt{x^2 + y^2}}$$
This is a correct rectangular form.
That's it. You are done! ;)
To make it a little easier, we can also do (using $(1)$):
\begin{array}{}
r&=&4+4\cos(\theta) \\
r&=&4+4\frac x r \\
r^2&=&4r + 4x \\
x^2+y^2&=&4\sqrt{x^2+y^2} + 4x
\end{array}

 

[soroban]

Hello, Raerin!

[tex]\text{Convert }\,r\:=\:4+4\cos\theta\,\text{ to rectangular form.}[/tex]

I would do it like this . . .[tex]\text{We have: }\:r \:=\:4(1 + \cos\theta)[/tex]

[tex]\text{Multiply by }r\!:\;r^2 \:=\:4(r + r\cos\theta)[/tex]

[tex]\text{Convert: }\:x^2+y^2 \:=\:4\left(\sqrt{x^2+y^2} + x\right)[/tex]