How do you convert #(x-3)^2+(y+1)^2=10# to polar form?

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Answer 1 · 2016-03-10T23:15:27

#r=6 # cos # \theta - 2# sin #\theta#

First use the binomial theorem to express the squared terms as polynomials. Recall that #(a+b)^2=a^2+2ab+b^2#. So

#(x-3)^2=x^2-6x+9#
#(y+1)^2=y^2+2y+1#

Now write the left side ss the sum of these polynomials. You should put #x^2+y^2# together because we know that in polar form that is #r^2#.

#(x^2+y^2)-6x+2y+10=10#

Now put in the coordinate conversions

#x=r # cos #\theta#
#y=r # sin #\theta#
#x^2+y^2=r^2#

Then

#r^2-6r# cos #\theta+2r# sin #\theta=0#

This can be factored so we have two possibilities:

#r=0#

#r-6 # cos #\theta+2# sin #\theta=0#

The first equation is just the origin and the second equation already has a point at the origin. So we just use the second equation.