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How do you convert #x² + (y + 4)² = 16# to polar form?

1 Answer

Shwetank Mauria

Oct 7, 2016

#r^2+8rsintheta=0#

Explanation:

The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is #x=rcostheta# and #y=rsintheta#.

Therefore ## can be written as

#(rcostheta)^2+(rsintheta+4)^2=16#

or #r^2cos^2theta+r^2sin^2theta+8rsintheta+16=16#

or #r^2(cos^2theta+sin^2theta)+8rsintheta=0#

or #r^2+8rsintheta=0#

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