$\begingroup$

I have the equation $(x-3)^2+y^2=9$ which is a circle centered at 3 and I need to find the polar coordinates for this equation, but I am not sure where to start because the $(x-3)$ is throwing me off. Can I start doing $(r\cos(\theta)-3)^2+r\sin(\theta)^2=9$? Where do I go from there? I am pretty thrown off, but I do know that $r^2 = x^2+y^2$. How could I get it in that form? Once there how would I find $\theta$?

$\endgroup$

2 Answers

$\begingroup$

You may write $(x-3)^2+y^2=9$ as $$ \begin{align} (r \cos \theta-3)^2+(r \sin \theta)^2&=9 \\\\ r^2 \cos^2 \theta+r^2 \sin^2 \theta-6r \cos \theta+9&=9 \\\\ r^2 (\cos^2 \theta+\sin^2 \theta)-6r \cos \theta&=0 \\\\ \cos \theta&=\frac{r}6. \end{align} $$

$\endgroup$ 1 $\begingroup$

$ \theta = \cos^ {-1} (r/6), r =0. $ It is a circle through center, centered on x-axis, diameter 6.

$\endgroup$ 5

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy