jmc

We try completing the square in $x$ again, that gives $$(x-3{)}^2-9+2y^2-8y+21=0.$$Then completing the square in $y$ gives $$(x-3{)}^2-9+2(y-2{)}^2-8+21=0.$$Combining all the constants we have $$(x-3{)}^2+2(y-2{)}^2=-4.$$The left hand side is always nonnegative, so this graph is $\boxed{\text{empty}}$.