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

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Answer 1 · 2015-11-28T09:43:15

$r = 4 cos (θ)$ $r = 4cos(theta)$

Using the equations

${(x = rcos(theta)), (y=rsin(theta)):}$

(for an explanation as to where those equations come from, see How do you convert rectangular coordinates to polar coordinates? )

we can substitute in to obtain

$(rcos(theta)-2)^2 + (rsin(theta))^2 = 4$

$=> r^2cos^2(theta)-4rcos(theta) + 4 + r^2sin^2(theta) = 4$

$=> r^2(cos^2(theta)+sin^2(theta)) - 4rcos(theta) = 0$

$=> r^2 - 4rcos(theta) = 0$

$=> r = 4cos(theta)$

How do you convert (x - 2)^2 + y^2 = 4(x−2)2+y2=4(x - 2)^2 + y^2 = 4 in polar form?