Find the flux of $F=(x-y)i+xj $ across the circle $x^{2}+y^{2}=1$ in the $xy$-plane.
Find the circulation of the field $F=(x-y)i+xj$ around the circle $r(t)=(cos (t))i +(sin(t)) j,0\leq t \leq 2 \pi.$
I think both are same. Both can be solved by Green's theorem.
The flux of a vector field $F$ across some boundary $S$ is, if we think of the vector field as describing the movement of some fluid, the volume of that fluid crossing $S$ in a given unit time. It is given, in 2 dimensions, by the flux integral $$\int_SF\cdot \mathbf{n}ds,$$
where $\mathbf{n}: (x,y)\mapsto \mathbf{n}(x,y)$ is a function defined on $S$ at gives the (outward) unit normal vector to $S$ for any $(x,y)$ in $S$.
In this case, $\mathbf{n}(x,y) = (x,y)$, so our integral is just
$$\int_S(x-y)x + xyds = \int_Sx^2ds.$$
To calculate this, we parameterise the circle using your parameterisation from your second question, so $x = \cos(t)$, $y = \sin(t)$, and $ds = \|(-\sin(t),\cos(t)\|dt = dt$, so our integral is
$$\int\limits_0^{2\pi}cos^2(t)dt = \pi.$$
For your second question, however, matters are different: Intuitively, the circulation is how much of the vector field is going around the circle, rather than across it. It is given by
$$\int_SF\cdot ds = \int_SF(s)\cdot s'(t)dt.$$
In this case, we again use the same parameterisation of our circle, so our integral is
$$\int\limits_0^{2\pi}\left(\array{\cos(t)-\sin(t)\\\cos(t)}\right)\cdot\left(\array{-\sin(t)\\\cos(t)}\right)dt = \int\limits_0^{2\pi}1-\cos(t)\sin(t)dt = 2\pi.$$
So no, the questions are not both the same.