Consider $\mathbb{R}^2$ equipped with the $p$-norm \begin{eqnarray} \Vert (x_1,x_2) \Vert_p:=(|x_1|^p+|x_2|^p)^\frac{1}{p} \end{eqnarray} where $1<p<\infty$. For $x\in \mathbb{R}^2$ and $l \geq 0$ let $S_l(x):=\{y\in\mathbb{R}^2 \mid \Vert x-y\Vert_p=l\}$ be the sphere of radius $l$ around $x$. It is well-known (see e.g. here) that in the case where $p=2$ the sphere/circle is uniquely determined by three points on it, meaning that for distinct points $y_1,y_2,y_3 \in \mathbb{R}^2$ there exists at most one $x\in\mathbb{R}^2$ and $l\geq0$ such that $y_1, y_2, y_3 \in S_l(x)$. I was wondering if that is true for other $p$ as well. I expect this to hold but I do not see the proof.
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