How do you write the equation using polar coordinates given #(x-2)^2+y^2=4#?

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Answer 1 · 2016-12-01T11:05:29

This is the equation of a circle with center in (2,0) and radius 2 and becomes:

#rho^2 -4rho cos theta =0#

Substitute the expression of #x# and #y# in terms of polar coordinates:

#x=rho cos theta#
#y=rho sin theta#

and you have:

#(rho cos theta - 2)^2+rho^2 sin^2 theta = 4#

#rho^2cos^2theta -4rho cos theta +cancel 4 + rho^2 sin^2 theta = cancel 4#

#rho^2(cos^2theta +sin^2 theta) -4rho cos theta =0#

#rho^2 -4rho cos theta =0#