How do you convert #(x-3)^2+(y-2)^2=13# to polar form?

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Answer 1 · 2016-06-22T14:46:20

#r=3 cos theta + 2 sin theta#.

Use the transformation #(x, y)=(r cos theta, r sin theta)#.

The given equation reduces to the polar form

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

So, #r = 0 and r=2(3 cos theta + 2 sin theta)#. The second equation

includes the first at #(0, arc tan (-3/2))#, obtained by solving

. #3 cos theta + 2 sin theta=0#.

The circle passes through the pole.

The radius of the circle is #sqrt 13#.

Note that arc tan (-3/2) has two values = #123.6^o and 303.69^o#,

nearly, in #(0, 360^o)#, giving the opposite directions of the tangent

to the circle at the origin (pole)..

The center of the circle is at#(sqrt 13, arc tan (-3/2)-or+90^o)#.

I have taken the trouble of giving some details relating to the polar

form. I think that Just giving the polar equation alone, sans further

details, is an incomplete exercise..