Hi I need your help with this! What's the Cartesian equation for this: r = (3)+(-5)*COS(θ)

Question

amistre64 · Answer

x = r cos(theta) .... might be useful

amistre64 · Answer

r = (3)+(-5)*COS(θ)   multiply by r

r^2 = 3r -5 rCOS(θ)

r^2 = 3r - 5x  ;  r^2 = x^2 + y^2

x^2 + y^2 = 3r - 5x;  any way to get rid of that pesky r?

amistre64 · Answer

x^2 + y^2 = 3sqrt(x^2+y^2) - 5x

x^2 + 5x + y^2 = 3sqrt(x^2+y^2)

(x^2 + 5x + y^2)^2 = 9x^2 + 9y^2 ... maybe

amistre64 · Answer

http://www.wolframalpha.com/input/?i=%28x%5E2+%2B+5x+%2B+y%5E2%29%5E2+%3D+9x%5E2+%2B+9y%5E2

anonymous · Answer

Hey! I do not mean to sound ungrateful for not replying so I had to say this. THANK YOU. THANK YOU SO MUCH. Honestly you saved my life right there, I was so lost but when I read you answer I understood and thank you so much for the visual on the image. I thought it would look like that but I was unsure. You are really awesome amistre64! Again thank you.