How do you convert #(x-2)^2 + (y-3)^2 = 1# into a polar equation?

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How do you convert #(x-2)^2 + (y-3)^2 = 1# into a polar equation?

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Answer 1 · 2016-05-13T19:34:52

#r^2-2(2 cos(theta)+3 sin(theta))r+12=0#

Explanation:

The pass relationships #x=r cos(theta),y=r sin(theta)# substituted in the cartesian circunference representation, gives:
#(r cos(theta)-2)^2+(r sin(theta)-3)^2=1#. Simplifiying
#r^2-2(2 cos(theta)+3 sin(theta))r+12=0#

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How do you convert #(x-2)^2 + (y-3)^2 = 1# into a polar equation?

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