From the diagram we can see that point P has polar coordinates
#( r , theta )# and Cartesian coordinates #(x,y)#.
And #color(white)(88)x=rcos(theta)# , #y = rsin(theta)#
#(x,y) -> (rcos(theta), rsin(theta))#
Also:
By Pythagoras' Theorem :
#r^2=(rcostheta)^2+(rsintheta)^2#
Since:
#x=rcos(theta)# and #y = rsin(theta)#
Then:
#r^2=x^2+y^2# #:. r=sqrt(x^2+y^2)#
Using these ideas:
#r=2+cos(theta)#
Substitute: #color(white)(88)r=sqrt(x^2+y^2)#
#sqrt(x^2+y^2)=2+cos(theta)#
Substitute: #color(white)(88)cos(theta)=x/r#
#sqrt(x^2+y^2)=2+x/r#
Substitute fro #bbr# again:
#sqrt(x^2+y^2)=2+x/sqrt(x^2+y^2)#
Multiply by: #sqrt(x^2+y^2)#
#sqrt(x^2+y^2)*sqrt(x^2+y^2)=2sqrt(x^2+y^2)+(xsqrt(x^2+y^2))/sqrt(x^2+y^2)#
#x^2+y^2=2sqrt(x^2+y^2)+(xcancelsqrt(x^2+y^2))/cancelsqrt(x^2+y^2)#
#x^2+y^2=2sqrt(x^2+y^2)+x#
#x^2+y^2-x=2sqrt(x^2+y^2)#
Squaring:
#x^4+2x^2y^2-2x^3+y^4-2y^2x+x^2=4x^2+4y^2#
#color(blue)(x^4+2x^2y^2-2x^3+y^4-2y^2x+x^2-4x^2-4y^2=0)#
I did not include the steps in expanding #(x^2+y^2-x)^2#. This would make the answer too long.