Let $D\subset R^n$ be a smoothly bounded open domain and $0\in D$. For any $x\in\partial D$ there holds \begin{eqnarray*} 2 \,a'(\vert x\vert)\,(x\cdot\nu(x))+(n-1)\,a(\vert x\vert)\,H(x)=const.. \end{eqnarray*} Here $a:(0,\infty)\to (0,\infty)$ is a positive smooth function, $a'$ denotes its derivative, $\nu(x)$ is the outer unit normal vector in $x\in\partial D$ and $(n-1)H(x)$ denotes the mean curvature in $x\in\partial D$. Does this imply that $D$ is a ball?

Let $D\subset\mathbb R^n$ be a smoothly bounded open domain and $0\in D$. For any $x\in\partial D$ there holds \begin{eqnarray*} 2 \,a'(\vert x\vert)\,(x\cdot\nu(x))+(n-1)\,a(\vert x\vert) \, H(x) = \text{const.} \end{eqnarray*} Here $a:(0,\infty)\to (0,\infty)$ is a positive smooth function, $a'$ denotes its derivative, $\nu(x)$ is the outer unit normal vector in $x\in\partial D$ and $(n-1)H(x)$ denotes the mean curvature in $x\in\partial D$. Does this imply that $D$ is a ball?