transform the following polar equation into an equation in rectangular coordinates r=-4 sin theta would it be x^2(y+2)^2=4?

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transform the following polar equation into an equation in rectangular coordinates r=-4 sin theta would it be x^2(y+2)^2=4?

Mathematics
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\(r = -4\sin\theta\) is a circle
the graph wont change between polar and cartesian
does your equation look anything like the eqn of a circle ?

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Other answers:

no so y=-4?
aren't you just guessing the options ?
Remember \[x = r \cos \theta\]\[y= r \sin \theta\]\[r^2 = x^2+y^2 \implies r = \sqrt{x^2+y^2}\]
Multiply both sides by \(r\) to see it.
how would I plug in the numbers though?
\[r^2 = -4rsin \theta \] doing as parth suggested, see what you can do with this given the information above.
\[\color{red}{r^2} = -4\color{blue}{r\sin\theta}\] you don't want to see \(r, \theta\) so replace \(\color{red}{r^2} \) by \(\color{red}{x^2+y^2} \) and \(\color{blue}{r\sin\theta}\) by \(\color{blue}{y}\) just algebra circus
therefore X+Y=-4? I get it
Not quite..
\[x^2+y^2=-4y\] \[r^2 = x^2+y^2~~~~y = r \sin \theta\]
And then from there you just substitute right?
technically we're done with the transformation \(x^2+y^2=-4y\) is the corresponding rectangular equation are you given options or something ?
I was right actually because x^2(y+2)^2=4 was correct
you're correct upto a + sign
completing the square gives you \(x^2\color{red}{+}(y+2)^2=4\) which is slightly different from \(x^2(y+2)^2=4\) lexically, but totally different semantically. you might think "its just a +"... graph each of them and see
\[x^2+y^2=-4y\]\[x^2+(y^2+4y)=0\]\[x^2+(y^2+4y+\color{red}{4})=4\]\[\boxed{x^2+(y+2)^2 = 4}\]
|dw:1439362067882:dw|
the answer is right. The user just forgot to add the + sign during the final stages of this problem. \[x^2\color{red}{+}(y+2)^2=4 \]

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