Polar to Cartesian relations:
#r^2=x^2+y^2#
#theta=arctan(y/x)#
Substitute these in:
#4-r=sintheta-2cos^2theta#
#4-sqrt(x^2+y^2)=sinarctan(y/x)-2cos^2arctan(y/x)#
We can simplify the RHS; we need to express sine and cosine in terms of tangent.
Sine
#tanx=sinx/cosx=sinx/sqrt(1-sin^2x)#
#(1-sin^2x)tan^2x=sin^2x#
#tan^2x=sin^2x(1+tan^2x)#
#sinx=tanx/sqrt(1+tan^2x)#
So #sinarctanx=x/sqrt(1+x^2)#
Cosine
#tanx=sinx/cosx=sqrt(1-cos^2x)/cosx#
#cos^2xtan^2x=1-cos^2x#
#cos^2x(1+tan^2x)=1#
#cosx=1/sqrt(1+tan^2x)#
So #cosarctanx=1/sqrt(1+x^2)#
Substitute
So
#sinarctan(y/x)=(y/x)/sqrt(1+y^2/x^2)=y/sqrt(x^2+y^2)#
and
#cos^2arctan(y/x)=[1/sqrt(1+y^2/x^2)]^2=[x/sqrt(x^2+y^2)]^2=x^2/(x^2+y^2)#
We have after Polar to Cartesian substitution
#4-sqrt(x^2+y^2)=sinarctan(y/x)-2cos^2arctan(y/x)#
which then becomes
#4-sqrt(x^2+y^2)=y/sqrt(x^2+y^2)-(2x^2)/(x^2+y^2)#
Put over a common denominator:
#4(x^2+y^2)-(x^2+y^2)^(3/2)=ysqrt(x^2+y^2)-2x^2#
#6x^2+4y^2=(x^2+y+y^2)sqrt(x^2+y^2)#
We can square out the root to create a polynomial in #x# and #y#, which is a more conventional way to express this, but less tidy. It also risks creating surplus roots to the equation that weren't roots of the original.
#(6x^2+4y^2)^2=(x^2+y+y^2)^2(x^2+y^2)#
#36x^4+48x^2y^2+16y^2=(x^4+2x^2y+2x^2y^2+y^2+2y^3+y^4)(x^2+y^2)#
#36x^4+48x^2y^2+16y^2=
x^6+2x^4y+2x^4y^2+x^2y^2+2x^2y^3+x^2y^4+
x^4y^2+2x^2y^3+2x^2y^4+y^4+2y^5+y^6#
Collect terms and group them according to the degree of each monomial:
#(x^6+3x^4y^2+3x^2y^4+y^6)+(2x^4y+4x^2y^3+2y^5)+(-36x^4-47x^2y^2+y^4)+(-16y^2)=0#
Notice that we can factorise a couple of the groups, a process that does some returning along the path taken:
#(x^2+y^2)^3+2y(x^2+y^2)^2-36x^4-47x^2y^2+y^4-16y^2=0#
