How do you convert #r = 8 sin theta# into cartesian form?

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Answer 1 · 2016-12-29T08:58:00

The answer is #x^2+(y-4)^2=4^2#

To convert from polar coordinates #(r,theta)# to cartesian coordinates #(x,y)#, we use the following equations

#x=rcostheta#

#y=rsintheta#

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

Here, we have

#r=8sintheta#

#r=8*y/r#

#r^2=8y#

#x^2+y^2=8y#

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

Completing the squares

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

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

This is the equation of a circle, center #(0,4)# and radius #=4#

graph{x^2+(y-4)^2=16 [-13.88, 14.6, -2.85, 11.39]}

Answer 2 · 2016-12-29T09:05:20

#x^2+y^2=8y#

we understand that;

#x=r cos theta#,
#y=r sin theta# and
#x^2+y^2 = r^2# --@

from the question,

#r=8sin theta#
we can multiply with r, then

#r*r = r*8 sin theta#
#r^2= 8 (r sin theta) where, y = r sin theta#
#r^2 = 8y#

therefore from @ the answer is
#x^2+y^2=8y#