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

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To convert the polar equation r=4+4cos(θ) into rectangular form, start with the relationships r²=x²+y² and rcos(θ)=x. The initial equation can be rewritten as √(x²+y²) = 4 + 4(x/√(x²+y²)). This leads to the equation x²+y² = 4√(x²+y²) + 4x after squaring both sides. The conversion process is confirmed to be correct, resulting in a valid rectangular form.
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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.
 
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Raerin said:
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}
 
Hello, Raerin!

\text{Convert }\,r\:=\:4+4\cos\theta\,\text{ to rectangular form.}
I would do it like this . . .\text{We have: }\:r \:=\:4(1 + \cos\theta)

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

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