Find the polar equation of the circle with center on the line theta = pi, of radius 1, and passing through the origin.

Question

Question

Find the polar equation of the circle with center on the line theta = pi, of radius 1, and passing through the origin.

i set a point (a,pi) on the line and going to this equation

a = rsin(theta) but its wrong.

where did i am doing wrong ?

Answer 1

The points which correspond to $\theta = \pi$ are exactly those on the $x$-axis. If you want the circle to pass through the origin, the only way to do this is to have it being tangent to $(0,0)$ i.e it has center $(-1,0)$ or $(1,0)$. This equation is given by

$$r(\theta) = (\cos \theta \pm 1, \sin \theta); \ \theta \in [0,2\pi]$$

Answer 2

To find the polar equation of this circle, we can start with its cartesian equation $$(x+1)^2+y^2 = 1.$$ Expanding and rearranging this equation, we get $$x^2+2x+1+y^2 = 1 \\ x^2+y^2+2x = 0.$$ Finally, substitute $x=r\cos\theta$ and $y=r\sin\theta$ to make this $$r^2+2r\cos\theta=0.$$
This can also be derived somewhat indirectly by starting with the parametrization $x=\cos t-1$, $y=\sin t$. Note that the parameter $t$ is not the polar angle $\theta$.