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How do you convert # (r-1)^2= - sin theta costheta +costheta# to Cartesian form?

1 Answer

Shwetank Mauria

Nov 8, 2016

#x^2+y^2-2sqrt(x^2+y^2)+1=-(xy)/(x^2+y^2)+x/sqrt(x^2+y^2)#

Explanation:

The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is given by

#x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#

Hence, #(r-1)^2=-sinthetacostheta+costheta#

#hArrr^2-2r+1=-sinthetacostheta+costheta#

or #x^2+y^2-2sqrt(x^2+y^2)+1=-(y/r*x/r)+x/r#

or #x^2+y^2-2sqrt(x^2+y^2)+1=-(xy)/(x^2+y^2)+x/sqrt(x^2+y^2)#

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