## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$8x^{2}+9y^{2}-6x-9=0$
Multiply $3-\cos\theta$ to both sides pf the equation to obtain: \begin{align*} (3-\cos\theta)(r)&=\frac{3}{3-\cos\theta} \cdot (3-\cos\theta)\\ \\3r-r\cos\theta&=3 \end{align*} Since $r=\sqrt {x^{2}+y^{2}}$ and $r\cos\theta=x$, then the equation above is equivalent to \begin{align*} 3\left(\sqrt {x^{2}+y^{2}}\right)-x&=3\\ 3\left(\sqrt {x^{2}+y^{2}}\right)&=x+3\\ \left(3\left(\sqrt {x^{2}+y^{2}}\right)\right)^2&=(x+3)^2\\ 9(x^{2}+y^{2})&=x^2+6x+9\\ 9x^{2}+9y^{2}&=x^2+6x+9\\ 9x^{2}+9y^{2}-(x^2+6x+9)&=0\\ 9x^{2}+9y^{2}-x^2-6x-9&=0\\ 8x^{2}+9y^{2}-6x-9&=0\\ \end{align*}