Question

I understand how the cartesian equation was obtained, but I don't get how the result is obtained.

Answer 1

\[r = 8\sin(\theta) + 8\cos(\theta)\]
Square both sides of the equation, so that we get $$r^2 \to x^2 + y^2$$$$rsin(\theta) \to x $$and$$rcos(\theta) \to y$$
\[r^2 = 8rsin(\theta) + 8rcos(\theta)\]
\[x^2 + y^2 = 8y + 8x\]

But now, how do I go from this step to

\[(x-4)^2 + (y-4)^2 = 32\]

?

Answer 2

\[x^2-8x + y^2-8y=0\]
Then complete the sqaure for each polynomial
\[(x^2-8x+16)-16+(y^2-8y+16)-16=0\]
Can you see the final step? ^_^

Answer 3

Ahh! Yes.

Bring both -16's over to the right side of the equation, and since both of the polynomials are perfect squares, the whole equation condenses to:

\[x^2 -8x +y^2 -8y = 0\]
\[(x^2 -8x +16) -16 + (y^2 -8y +16) -16 = 0\]
\[(x^2 -8x +16)+ (y^2 -8y +16) = 32\]
\[(x -4)^2+ (y -4)^2 = 32\]

Thank you so much. I don't know why I had so much trouble seeing that.

Answer 4

yup yup! ^_^ And since you had already done the complicated part, you were probably thinking too hard!