How do you convert #r = 8sin(θ)# into rectangular form?

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Answer 1 · 2018-04-14T12:06:22

Rectangular form is #x^2+(y-4)^2= 4^2 #

The relation between polar and Cartesian coordinates are

#r^2=x^2+y^2 , tan theta =y/x , x=r cos theta , y=r sin theta #

#r= 8 sin theta :. r*r=r*8 sin theta # or

# r^2=8 r sin theta :. x^2+y^2= 8 y# or

#x^2+y^2- 8 y=0 or x^2+y^2- 8 y+16= 16 # or

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

Rectangular form is #x^2+(y-4)^2= 4^2 #

graph{x^2+(y-4)^2=16 [-20, 20, -10, 10]}