How do you convert #x^2+4x+y^2+4y=0# to polar form?

Question

4602 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-19T10:25:06

#r+4sqrt2cos(theta-pi/4)=0#

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

#x=rcostheta# and #y=rsintheta# or #r^2=x^2+y^2#

Hence #x^2+4x+y^2+4y=0# can be written as

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

or #r^2+4rcostheta+4rsintheta=0#

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

or #r+4(costheta+sintheta)=0#

or #r+4sqrt2(costhetaxx1/sqrt2+sinthetaxx1/sqrt2)=0#

or #r+4sqrt2(costhetacos(pi/4)+sinthetasin(pi/4))=0#

or #r+4sqrt2cos(theta-pi/4)=0#