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

1. ganeshie8

$$r = -4\sin\theta$$ is a circle

2. ganeshie8

the graph wont change between polar and cartesian

3. ganeshie8

does your equation look anything like the eqn of a circle ?

4. anonymous

no so y=-4?

5. ganeshie8

aren't you just guessing the options ?

6. anonymous

Remember $x = r \cos \theta$$y= r \sin \theta$$r^2 = x^2+y^2 \implies r = \sqrt{x^2+y^2}$

7. ParthKohli

Multiply both sides by $$r$$ to see it.

8. anonymous

how would I plug in the numbers though?

9. anonymous

$r^2 = -4rsin \theta$ doing as parth suggested, see what you can do with this given the information above.

10. ganeshie8

$\color{red}{r^2} = -4\color{blue}{r\sin\theta}$ 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

11. anonymous

therefore X+Y=-4? I get it

12. anonymous

Not quite..

13. anonymous

$x^2+y^2=-4y$ $r^2 = x^2+y^2~~~~y = r \sin \theta$

14. anonymous

And then from there you just substitute right?

15. ganeshie8

technically we're done with the transformation $$x^2+y^2=-4y$$ is the corresponding rectangular equation are you given options or something ?

16. anonymous

I was right actually because x^2(y+2)^2=4 was correct

17. ganeshie8

you're correct upto a + sign

18. ganeshie8

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

19. anonymous

$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}$

20. triciaal

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21. UsukiDoll

the answer is right. The user just forgot to add the + sign during the final stages of this problem. $x^2\color{red}{+}(y+2)^2=4$