At a certain stage of my research work, I require the following fact to hold true.


A surface $S$ satisfies certain conditions so that it lies on a 4-sphere in $R^{21}$ (I have used the results proved in <http://www.sciencedirect.com/science/article/pii/S0926224508001125>). Hence, it also lies on a (20-dimensional) hypersphere that passes through the origin, because of the following claim.


**Claim.** Let us consider $\mathbb{R}^n$ with a sufficiently large $n\in\mathbb{N}$. Any $p$-dimensional sphere embedded therein, with $p<n-1$, always lies on a $k$-dimensional sphere, for each $k=p+1,\ldots,n-1$, that passes through a given point $\mathbf{a}\in\mathbb{R}^n$ ($\mathbf{a}$ lies outside the ($p$+1)-plane containing the given $p$-sphere).


My question is, is my above claim true? I am not being able to prove it. Any help would be gratefully appreciated.



Do lower dimensional spheres always lie on higher dimensional spheres?
This is an exercise in analytic geometry.


Let $x$ be the center of the $p$-sphere, which is contained in the affine $(p+1)$-plane $x+V$, and let $b$ be any point on the sphere. 


Then we need to find a point $y$ in $x+V^\perp$ such that $\lvert y-a\rvert^2 = \lvert y-b\rvert^2$. This equation is equivalent to the fact that $y$ is an element of $$\frac12(a+b)+(a-b)^\perp.$$
Since $a-b$ is not in $V$, the hyperplane $(a-b)^\perp$ does not contain $V^\perp$. This means the intersection of $\frac12(a+b)+(a-b)^\perp$ and $x+V^\perp$is non-empty, so there does exist a solution $y$, which means there does always exist such an $(n-1)$-sphere.


Intersecting such an $(n-1)$-sphere with an affine $k$-space through $a$ and $p+2$ points of the $p$-sphere spanning $x+V$ (so that $k\ge p+2$) we obtain the desired $(k-1)$-sphere.


