How do you convert #r=4sin theta# into cartesian form?

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Answer 1 · 2017-02-04T14:24:03

#x^2+y^2-y=0#

for polar to and from Cartesian the equations are

#r^2=x^2+y^2#

#x=rcostheta#

#y=rsintheta#
#:.r=4sintheta " multiply through by "r#

#r^2=rsintheta#

substituting back using the above conversions

#x^2+y^2=4y#

#x^2+y^2-4y=0#

graph{x^2+y^2=4y [-10, 10, -5, 5]} this is a circle centre #(0,2) # radius #=2#