Compute $$\oint_C \mathbf{B} \cdot d\mathbf{r}$$ where $\mathbf{B} = \dfrac{1}{[(x-2)^2+(y-3)^2)]}[(y-3)\mathbf{e}_x-(x-2)\mathbf{e}_y] $ and $C: \: x^2+y^2 = 16$ with your choice of orientation.
Attemped solution
Let $$\begin{cases} u = x-2 \\v=y-3 \end{cases} $$ this reduces $\mathbf{B}$ to $\dfrac{1}{u^2+v^2}(v\mathbf{e}_u-u\mathbf{e}_v)$. I'm not sure how to proceed without introducing polar coordinates, is there any neat way to tidy this up?