Chapter 10 - Practice Exercises - Page 593: 31

center $(\sqrt 2,0)$ and radius $\sqrt 2$

Here, we have $r^2=2\sqrt 2 r \cos \theta$ This implies that $x^2+y^2-2\sqrt 2 x=0$ or, $(x-\sqrt 2)^2+y^2=2$ Thus, we have an equation of a circle with center $(\sqrt 2,0)$ and radius $\sqrt 2$