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

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Answer 1 · 2015-11-28T09:43:15

#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)#