How do you find the polar equation for x^2+(y-2)^2-4=0#?

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How do you find the polar equation for x^2+(y-2)^2-4=0#?

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Answer 1 · 2016-07-24T17:14:51

$r = 4 sin θ$ $r=4sintheta$

Explanation:

The relation between polar coordinates $(r, θ)$ $(r,theta)$ and rectangular coordinates $(x, y)$ $(x,y)$ are given by $x = r cos t ˆ a$ $x=rcosthata$ and $y = r sin θ$ $y=rsintheta$ and hence $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$

Hence $x^{2} + {(y - 2)}^{2} - 4 = 0$ $x^2+(y-2)^2-4=0$ can be written as

$x^{2} + y^{2} - 4 y + 4 - 4 = 0$ $x^2+y^2-4y+4-4=0$ or using relations between polar and Cartesian coordinates

$x^{2} + y^{2} - 4 y = 0$ $x^2+y^2-4y=0$

$r^{2} - 4 r sin θ = 0$ $r^2-4rsintheta=0$ or

$r - 4 sin θ = 0$ $r-4sintheta=0$ or $r = 4 sin θ$ $r=4sintheta$

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How do you find the polar equation for x^2+(y-2)^2-4=0#?

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