How do you convert $x^{2} + y^{2} - 2 y = 0$ $x^2+y^2 - 2y=0$ into polar form?

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How do you convert $x^{2} + y^{2} - 2 y = 0$ $x^2+y^2 - 2y=0$ into polar form?

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Answer 1 · 2016-07-13T01:15:21

Make use of a few conversion formulas and simplify. See below.

Explanation:

Recall the following formulas, used for conversion between polar and rectangular coordinates:

$x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$
$r sin θ = y$ $rsintheta=y$

Now take a look at the equation:
$x^{2} + y^{2} - 2 y = 0$ $x^2+y^2-2y=0$

Since $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$ , we can replace the $x^{2} + y^{2}$ $x^2+y^2$ in our equation with $r^{2}$ $r^2$ :
$x^{2} + y^{2} - 2 y = 0$ $x^2+y^2-2y=0$
$\to r^{2} - 2 y = 0$ $->r^2-2y=0$

Also, because $y = r sin θ$ $y=rsintheta$ , we can replace the $y$ $y$ in our equation with $sin θ$ $sintheta$ :
$r^{2} - 2 y = 0$ $r^2-2y=0$
$\to r^{2} - 2 (r sin θ) = 0$ $->r^2-2(rsintheta)=0$

We can add $2 r sin θ$ $2rsintheta$ to both sides:
$r^{2} - 2 (r sin θ) = 0$ $r^2-2(rsintheta)=0$
$\to r^{2} = 2 r sin θ$ $->r^2=2rsintheta$

And we can finish by dividing by $r$ $r$ :
$r^{2} = 2 r sin θ$ $r^2=2rsintheta$
$\to r = 2 sin θ$ $->r=2sintheta$

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How do you convert $x^{2} + y^{2} - 2 y = 0$ $x^2+y^2 - 2y=0$ into polar form?

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