Question

Find the Cartesian equation for r=2cos(theta) to identify the curve.

Answer 1

The most important two equations to know are:
\(\color{#000000}{ \displaystyle y=r\sin \theta }\)
\(\color{#000000}{ \displaystyle y=r\cos \theta }\)
-------------------------------------------
This is where they come from, basically.

\(\color{#ff0000 }{ \small ({\rm r},\angle \theta) }\)
\(\color{#ff0000}{{\tiny~~} _{■ }}\) \(\color{blue}{{\tiny~}\Huge _\text{─}}\)
\(\color{#ff0000}{{\tiny~} _{*}^{~~~*}}\)\({\small~~}\color{black}{ | }\) \(\color{blue}{\bf |}\)
\(\color{#ff0000}{{\tiny~} _{*}^{~~~*}}\) \(\quad \color{black}{ | }\) \(\color{blue}{\bf |}\)
\(\color{#ff0000}{r~~~{\tiny~} _{*}^{~~~*}}\) \(\quad \color{black}{ | }\) \(\color{blue}{{\tiny~}{ \color{red}{y} }}\)
\(\color{#ff0000}{{\tiny~} _{*}^{~~~*}}\) \(\quad \color{black}{ | }\) \(\color{blue}{\bf |}\)
\(\color{#ff0000}{_{*}^{~~~*~~~~~}\theta }\) \(\quad\color{black}{ | }\) \(\color{blue}{\bf |}\)
\(\color{#ff0000 }{ \small (0,0) }\) \(\color{#ff0000}{}\Huge ^{\large ^{\color{red}{ _■}}}\Huge ^{\text{──────}}\) \(\color{blue}{ \Huge ^-}\)
\(\color{blue}{ {{\bf |\text{─────}{{\tiny}}} \color{red}{x} }{\tiny}\text{─────}|}\)

\(\color{#000000}{ \displaystyle \sin \theta=y/r\quad \Longrightarrow \quad y=r\sin \theta }\)
\(\color{#000000}{ \displaystyle \cos \theta=x/r\quad \Longrightarrow \quad x=r\cos \theta }\)
-------------------------------------------
Then, using these two equations on (for example)
a general case - \(\color{#000000}{ \displaystyle r=b\cos\theta }\)

\(\color{#000000}{ \displaystyle r=b\cos\theta }\)
\(\color{#000000}{ \displaystyle r^2=br\cos\theta }\)

Note that
\(\color{#000000}{ \displaystyle x^2+y^2=\{r\cos\theta\}^2+\{r\sin \theta \}^2 }\)
\(\color{#000000}{ \displaystyle x^2+y^2=r^2\cos^2\theta+r^2\sin ^2\theta }\)
\(\color{#000000}{ \displaystyle x^2+y^2=r^2(\cos^2\theta+\sin ^2\theta) }\)
\(\color{#000000}{ \displaystyle x^2+y^2=r^2(1) }\)
\(\color{#000000}{ \displaystyle x^2+y^2=r^2 }\)

and recall that
\(\color{#000000}{ \displaystyle x=r\cos\theta }\)
and therefore,
\(\color{#000000}{ \displaystyle bx=br\cos\theta }\)

So, you get,
\(\color{#000000}{ \displaystyle r^2=br\cos\theta \quad \Longrightarrow \quad x^2+y^2=bx }\)

Answer 2

and then, by subtracting bx from both sides, and by completing the square you can identify what radius this circle will have (in my case it will depend on b), and where it is centered....