Convert $(x-3)^2 + y^2 = 49$ to polar form.

Question

Question: Convert $(x-3)^2 + y^2 = 49$ to polar form.

Answer: Applying $x=r\cos(\theta)$ and $y=r\sin(\theta)$, I find

$$x^2 - 6x + 9 + y^2 = 49$$

$$r^2-6x = 40$$

$$r^2-6r\cos(\theta) = 40$$

$$r\big(r-6\cos(\theta)\big) = 40$$

My answer appears to be wrong. What am I doing wrong?

Answer 1

That's correct indeed we have

$$x^2 - 6x + 9 + y^2 = 49 \implies r^2-6r\cos \theta=40$$

that is the polar form of a circle centered at $(3,0)$ and radius $7$.

If we refer the origin at $(3,0)$ we obtain indeed $r=7$.

Answer 2

To convert to polar form, set x=rcos($\theta$) and $y=rsin(\theta)=$ So we get $(rcos(\theta)-3)^2$+$r^2sin^2(\theta)=49$, expanding gives you $r^2cos(\theta)^2-6rcos(\theta)+9+r^2sin^2(\theta)=49$ then using the trig identity $sin^2(\theta)+cos^2(\theta)=1$ yields $r^2-6rcos(\theta)=40$