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So far I got

\begin{align} r & = 7 / (4 - 2 \cos\theta) \\ r (4 & - 2\cos\theta) = 7 \\ r (4 & - 2( x / r ) ) = 7 \end{align}

I apologize in advance for the bad formatting.

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You are on the right track. So we get $$4r-2x=7.$$ This implies that $$4r=7+2x.$$ Taking square on both sides, we get $$16r^2=(7+2x)^2.$$ Now note that $x^2+y^2=r^2$. Put this into the last equation, we have $$16(x^2+y^2)=(7+2x)^2.$$

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Thanks Paul. Is it appropriate to ask questions like this on here? I am new and here to learn. Thanks –  user1530249 Feb 5 '13 at 6:53
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Yes, here you can ask any question related to math. And it's good to let people know where you are stuck and what you don't understand, as you have already done. But you may want to learn how to type the equation using TeX. See here: meta.math.stackexchange.com/questions/107/… –  Paul Feb 5 '13 at 6:57
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