How do you convert #r = 7(1 - sin(θ))# into rectangular form?

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Answer 1 · 2017-12-17T17:32:13

#(x^2+7y+y^2)^2=49x^2+49y^2#

#r=7*(1-sin(theta))#

#r=7-7sin(theta)#

#r^2=7r-7rsin(theta)#

After using #r=sqrt(x^2+y^2)#, #r^2=x^2+y^2# and #y=rsin(theta)# inverse transforms,

#x^2+y^2=7sqrt(x^2+y^2)-7y#

#x^2+7y+y^2=7sqrt(x^2+y^2)#

#(x^2+7y+y^2)^2=49x^2+49y^2#