Chapter 11: Parametric Equations and Polar Coordinates - Section 11.7 - Conics in Polar Coordinates - Exercises 11.7 - Page 686: 55

Center at: $\quad (1,\pi)$ Radius $=1$

For $\displaystyle \theta=\frac{\pi}{2},\ r=0$, so the origin is on the circle. A circle passing through the origin, of radius $a$, centered at $P_{0}(r_{0}, \theta_{0}),$ has the polar equation $r=2a\cos(\theta-\theta_{0})$ Apply $\cos(\theta-\pi)=-\cos\theta$. We can write this equation as $ r=2(1)\cos(\theta-\pi),\qquad a=1, \theta_{0}=\pi$ Center at $\quad (1,\pi)$ Radius $=1$