How do you convert (r-1)^2= - sin theta costheta +costheta(r−1)2=−sinθcosθ+cosθ 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)x2+y2−2√x2+y2+1=−xyx2+y2+x√x2+y2

Explanation:

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

x=rcosthetax=rcosθ, y=rsinthetay=rsinθ and r^2=x^2+y^2r2=x2+y2

Hence, (r-1)^2=-sinthetacostheta+costheta(r−1)2=−sinθcosθ+cosθ

hArrr^2-2r+1=-sinthetacostheta+costheta⇔r2−2r+1=−sinθcosθ+cosθ

or x^2+y^2-2sqrt(x^2+y^2)+1=-(y/r*x/r)+x/rx2+y2−2√x2+y2+1=−(yr⋅xr)+xr

or x^2+y^2-2sqrt(x^2+y^2)+1=-(xy)/(x^2+y^2)+x/sqrt(x^2+y^2)x2+y2−2√x2+y2+1=−xyx2+y2+x√x2+y2

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