## Thomas' Calculus 13th Edition

A circle with center $(\sqrt 2,0)$ and radius $\sqrt 2$
Consider $r^2=2\sqrt 2 r \cos \theta$ or, $x^2+y^2-(2\sqrt 2) x=0$ This can be re-arranged as: $(x-\sqrt 2)^2+y^2=2$ Hence, we get the equation of a circle with center $(\sqrt 2,0)$ and radius $\sqrt 2$