Let $0\neq a\in R^n$ with $||a|| > 1$ . Let $S_1 (a)$ be the unit sphere around $a$. I'm looking for the points $c\in S_1 (a)$ s.t. the hyperplane tangent to $S_1 (a)$ at $c$ passes through the origin. (I think that in calculus it is called the affine tangent space to the manifold $S_1 (a)$ at $c$). When trying to plot it in $\mathbb{R}^2, \mathbb{R}^3$ I can see the problem has two solutions. I am looking for a general solution (for every $n$).
What I've tried: I can write $S_1 (a) = f^{-1}(1)$ where $f(x)= ||x-a||^{2}$, then $Df_{x} = 2(x-a)^{T}$ and for all $x\in S_1 (a)$ we have $Df_x \neq 0$ hence $T_p S_1 (a) = \ker Df_p$ . Thus the affine tangent space at c is $$T'_c S_1 (a) = \{c + y : (c-a)^T \cdot y = 0\}$$ Also I know that $||c-a|| = 1$. From the first equation, forcing $0\in T'_c S_1 (a)$ we have $0 = c+y \implies y=-c$ for some $y$ with $(c-a)^T \cdot y = 0 $ so $(c-a)^T \cdot c = 0$ but I guess this is just two equations with $n$ unknowns $c_1, ..., c_n$. what now?
Edit: (clarification) I am trying to obtain a closed formula for $c$ given a fixed $a$. If such formula exists, that is great. If there is a good approximation or an iterative method to converge to such $c$ it will also do the job.
Take $c=(c_1,c_2,\ldots,c_n)\in S_1(a)$ and let $a=(a_1,a_2,\ldots,a_n)$. Then\begin{align}f(x_1,x_2,\ldots,x_n)&=\|(x_1,x_2,\ldots,x_2)-(a_1,a_2,\ldots,a_n)\|^2\\&=(x_1-a_1)^2+\cdots+(x_n-a_n)^2.\end{align}So,$$\nabla f(c)=2(c_1-a_1,c_2-a_2,\ldots,c_n-a_n).$$It follows from this that the tangent plane of $S_1(a)$ at $c$ is the plane $P(c,a)$ defined by$$(c_1-a_1)(x_1-c_1)+(c_2-a_2)(x_2-c_2)+\cdots+(c_n-a_n)(x_n-c_n)=0,$$since it is orthogonal to $\nabla f(p)$ and $c\in P(c,a)$. Asserting that the origin belongs to $P(c,a)$ is the same thing as asserting that$$c_1(c_1-a_1)+c_2(c_2-a_2)+\cdots+c_n(c_n-a_n)=0$$or, which amounts to the same thing, that $\langle c,c-a\rangle=0$.