Polar coordinate form of circle equation

Question

I am trying to convert the equation $$(x-0.9)^2+y^2 = 0.1^2$$ into polar coordinates. Using $x = r\cos \theta$ and $y = r\sin \theta$ I get that

$$ r = \frac{1.8 \cos\theta \pm \sqrt{(1.8\cos\theta)^2-4\times0.8}}{2}$$

However I am not sure how to get my bounds for theta. What would they be in this situation?

Answer 1

I have another way to do it. Put $x = 0.9 + 0.1\cos \theta, y = 0.1\sin \theta $. In the equation, $r = 0.1$ already, and $\theta \in [0,2\pi]$

Answer 2

Your equation describes the circle with radius $0.1$ around the center $(0.9, 0)$.

We can take from the image that $\theta \in [-\theta_0, \theta_0]$ with $$ \tan(\theta_0) = \frac{0.1}{0.9} \iff \\ \theta_0 = \arctan(1/9) = 6.34^\circ $$

Answer 3

That is correct.

Around x-axis you get two values for a single $\theta$ , so there is no way you can avoid the $\pm$ sign, implying (+,-) for outside and inside respectively.

Best to always use parametric form as suggested by Kf-Sansoo and avoid the ambiguity.