Convert x^2+y^2=4y-2x to Polar Equation

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SUMMARY

The discussion focuses on converting the Cartesian equation x² + y² = 4y - 2x into polar coordinates. The initial substitution used is r² = 4y - 2x, with the polar transformations x = rcos(θ) and y = rsin(θ). The correct transformation leads to the equation r² = 4rsin(θ) - 2rcos(θ). Participants emphasize the importance of not taking the square root of both sides and caution against dividing by r, as it may lead to losing solutions when r equals zero.

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Homework Statement


Establish an equation in polar coordinates for the curve x^2+y^2=4y-2x


Homework Equations



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The Attempt at a Solution


I know that x^2+y^2=r^2 so I used substitution, and now have r^2=4y-2x. Now this next part, I'm really not sure if I'm allowed to do this... i know that x=rcos@ (let @=theta) and y=rsin@. So i simply substituted to get r^2=4rsin@-2cos@. Is that allowed? Also, i feel like my answer should just look like r=... So should I take the sqaure root of both sides?
Thank you, very appreciated.
 
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You should have r^2 = 4rsin@ - 2rcos@. You lost an "r" on your cosine term. You can simplify by dividing both sides by r, but you don't want to take the square root of both sides. You normally have to be careful of dividing by r, in case r happens to be equal to zero. That's not a problem in this case, since the curve doesn't go through the origin.
 
Yeah, the missing r is a type. Dividing makes sense though, duh. thanks so much!
 

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