How do you convert #1=(-x+4)^2+(2y+9)^2# into polar form? Trigonometry The Polar System Converting Between Systems 1 Answer Shiva Prakash M V Mar 1, 2018 #r^2(cos^2theta+4sin^2theta)-r(4costheta-18sintheta)+40=0# Explanation: Given: #1=(-x+4)^2+(2y+9)^2# #1=(-x+4)^2+(2(y+9/2))^2# #1=(-x+4)^2+4(y+9/2)^2# #x=rcostheta# #y=rsintheta# Substituting #1=(-rcostheta+4)^2+(2rsintheta+9)^2# #1=r^2cos^2theta-8rcostheta+16+4r^2sin^2theta+36rsintheta+81# #r^2(cos^2theta+4sin^2theta)+r(-8costheta+36sintheta)+81-1=0# #2r^2(cos^2theta+4sin^2theta)-2r(4costheta-18sintheta)+2(40)=0# #r^2(cos^2theta+4sin^2theta)-r(4costheta-18sintheta)+40=0# Answer link Related questions How do you convert rectangular coordinates to polar coordinates? When is it easier to use the polar form of an equation or a rectangular form of an equation? How do you write #r = 4 \cos \theta # into rectangular form? What is the rectangular form of #r = 3 \csc \theta #? What is the polar form of # x^2 + y^2 = 2x#? How do you convert #r \sin^2 \theta =3 \cos \theta# into rectangular form? How do you convert from 300 degrees to radians? How do you convert the polar equation #10 sin(θ)# to the rectangular form? How do you convert the rectangular equation to polar form x=4? How do you find the cartesian graph of #r cos(θ) = 9#? See all questions in Converting Between Systems Impact of this question 1948 views around the world You can reuse this answer Creative Commons License