How do you convert #4=(x+8)^2+(y+2)^2# into polar form?

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Answer 1 · 2017-01-06T07:12:43

#r^2+4r(costheta+sintheta)+64=0#

The relation between polar coordinates #(r,theta)# and corresponding Cartesian coordinates #(x,y)# is given by

#x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#.

Hence, #4=(x+8)^2+(y+2)^2# can be written as

#4=(rcostheta+8)^2+(rsintheta+2)^2#

or #r^2cos^2theta+16rcostheta+64+r^2sin^2theta+4rsintheta+4=4#

or #r^2(cos^2theta+sin^2theta)+16rcostheta+4rsintheta+64=0#

or #r^2+4r(costheta+sintheta)+64=0#