How do you convert #r = 4costheta - 4sintheta# to rectangular form?

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Answer 1 · 2016-10-06T14:00:58

Is a circle with origin at

#(2,-2)# and radius #2sqrt(2)#

Using the pass equations

#{(x=rcostheta),(y=rsintheta):}#

#r = 4 x/r-4 y/r# then

#r^2=4(x-y) = x^2+y^2# or

#x^2-4x+y^2+4y=0# which is a circle with origin at

#(2,-2)# and radius #2sqrt(2)#