see the bottom part of my reply to
http://www.jiskha.com/display.cgi?id=1269266887
How do I convert the polar equation r - 3cos(theta) = 5sin to cartesian equation form?
2 answers
For your equation
r - 3cosØ = 5sinØ
√(x^2 + y^2) - 3x/r = 5y/r
√(x^2 + y^2) = 5y/r + 3x/r
√(x^2 + y^2) = (5y + 3x)/r
√(x^2 + y^2) = (5y + 3x)/√(x^2 + y^2)
cross multiply
x^2 + y^2 = 5y + 3x
which looks like the equation of a circle to me
r - 3cosØ = 5sinØ
√(x^2 + y^2) - 3x/r = 5y/r
√(x^2 + y^2) = 5y/r + 3x/r
√(x^2 + y^2) = (5y + 3x)/r
√(x^2 + y^2) = (5y + 3x)/√(x^2 + y^2)
cross multiply
x^2 + y^2 = 5y + 3x
which looks like the equation of a circle to me
2 answers