How do you convert $x^{2} + {(y - 4)}^{2} = 16$ $x^2+(y-4)^2=16$ into polar form?

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Answer 1 · 2017-09-11T20:02:09

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Use the formulas $r^{2} = x^{2} + y^{2}, x = r cos θ, y = r sin θ$ $r^2=x^2+y^2, x= r cos theta, y= r sin theta$

$x^{2} + {(y - 4)}^{2} = 16$ $x^2+(y-4)^2 = 16$ -->FOIL

$x^{2} + y^{2} - 8 y + 16 = 16$ $x^2+y^2-8y+16=16$

$x^{2} + y^{2} - 8 y = 16 - 16$ $x^2+y^2-8y=16-16$

$x^{2} + y^{2} - 8 y = 0$ $x^2+y^2-8y=0$

$(x^{2} + y^{2}) - 8 y = 0$ $(x^2+y^2)-8y=0$

$r^{2} - 8 r sin θ = 0$ $r^2-8rsin theta=0$

How do you convert x^2+(y-4)^2=16x2+(y−4)2=16x^2+(y-4)^2=16 into polar form?