The parametric equations of a circle $C$ are: \begin{align*} x&=2+\dfrac{13}{5\sqrt{2}}\cos t\\ y&=1+\dfrac{13}{5\sqrt{2}}\sin t \end{align*} for $t\in[0,2\pi]$.
I am stuck on this part: Find the gradient function of the circle $C$, leaving it in terms of the parameter $t$.
The implicit equation of the given circle is $F(x,y)=(x-2)^2+(y-1)^2=R^2$, $R=13/5\sqrt{2}$. The gradient of the function $F$ is the vector field:
$$\mbox{grad}(F)=\left(\displaystyle\frac{\partial F}{\partial x}, \displaystyle\frac{\partial F}{\partial y}\right)^T=(2(x-2),\: 2(y-1))^T$$
Now you have to evaluate the gradient at the circle points:
$$\mbox{grad}(F)(x(t), y(t))=\left((26/5\sqrt{2})\cos{t},\: (26/5\sqrt{2})\sin{t}\right)^T$$