I am trying to understand the following claim.
Claim. Let $a_1, ..., a_n$ be $n$ points in $\mathbb R^d$, $d>n\geq 2$. We assume that $\{a_1, ..., a_n\}$ is linearly independent. Let $y$ be a point in the linear span of $a_1,..., a_n$ such that $$\langle y-a_i, a_i \rangle = 0 \text{ for all } i=1,..., n. \tag{1}$$ Then the ball with diameter joining $y$ and $0_{\mathbb R^d}$ is the $\color{red}{\text{smallest}}$ ball containing $\{a_1, ..., a_n\} \cup \{0_{\mathbb R^d}\}$.
My attempts. First, the existence of $y$ is well defined. Indeed, if we let $y=\sum_{i=1}^n x_i a_i$, then the system of linear equations $(1)$ has a unique solution $x=(x_i)_{i=1,...,n}$ by applying the Gram determinant. Second, let $v_0=\frac{y}{2}$, then I can prove that $||y-v_0||=||0_{\mathbb R^d} -v_0||=||a_i-v_0||$ for all $i=1,..., n$, i.e. the ball with diameter joining $y$ and $0_{\mathbb R^d}$ already contained $\{a_1, ..., a_n\} \cup \{0_{\mathbb R^d}\}$.
My question. To prove the smallestness, we may consider $B(v_1, R)$ the smallest ball containing $\{a_1, ..., a_n\} \cup \{0_{\mathbb R^d}\}$ so as to $v_1\neq \frac{y}{2}$. How do I deduce a contradiction?