How do you convert 1=(-x+4)^2+(2y+9)^2 into polar form?

1 Answer

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