Neumann data must integrate to zero

Let B(0,2) = {$(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} < 4$}, and $\partial{}B$ denote the boundary of B(0,2). Assume $(\alpha, \beta) \ne (0,0), k \in \mathbb{R}$, and u is any solution to $-\Delta{}u = 0$ in $B(0,2), \alpha{}u(x, y) + \beta(\partial{}u/\partial\nu)(x, y) = 1 + (x^{2} + y^{2})k$ on $\partial{}B$, where $\nu(x, y)$ is the unit outward normal to B(0,2) at $(x, y) \in \partial{}B$. Consider the following statements: $S_{1}$: If $\beta = 0$, then there exists $a (x_{0}, y_{0}) \in B(0,2)$ such that |$u(x_{0}, y_{0})$| = |1 + 4k|/|$\alpha$|. $S_{2}$: If $\alpha = 0$, then k = −1/4. Then