Answer
A circle with center $(\sqrt 2,0)$ and radius $\sqrt 2$
Work Step by Step
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 $\sqrt 2$
