# Trig polar equation problem

Had a trig exam today, got all problems right except for one that seemed to stump me:

Change from polar form to rectangular coordinate form:

r = 1 - 2cosθ

I got the graph right I know that, but I couldnt figure a way to change it over. It kind of bugs me because I went through my entire book after the exam and could not find one problem which had the same format, and no questions similar to this one were on the professors exam study guide. Hate it when professors do that. I think I got some crazy answer like (x^2+y^2)^2 = 5x^2 + y^2 or something like that. Pretty sure it is wrong though. Luckily, it was only 5pts out of 110pt test. Thanks in advanced for the replies.

Remember:

r^2 = x^2 + y^2
x = rcosθ
y = rsinθ
tanθ = y/x

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Try multiplying both sides of the equation by r and substitute directly from the conversion equations. Remember, r = sqrt(x^2 + y^2).

I got an answer different than yours when I did this.

By the way, was the shape of your graph a limaçon?

Try multiplying both sides of the equation by r and substitute directly from the conversion equations. Remember, r = sqrt(x^2 + y^2).

I got an answer different than yours when I did this.

By the way, was the shape of your graph a limaçon?
yea you are right, limacon

Ah ok, i did get the answer wrong, but I was pretty close! I'm sure he will give me partial credit for it because I was not too far off.

Thanks