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Earlier I asked a question on whether it is possible to find a sphere passing through a circle and a point non-coplanar to it. I wanted to know whether this was possible to do in higher dimensions.

In $\mathbb{R}^{k+2}$, given a $k$-sphere and a point outside the $k$-hyperplane containing it, can I find a $(k+1)$-sphere containing them?

While the geometric constructions described in the answers to the previous question seem to go through, I am not sure if they would hold in the higher dimensions or does the same intuition fail?

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1 Answers1

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Use the algebric approach:

Equation of the hypersphere of dimension $k$:

$\{(x_1,...,x_{k+2}) | \sum_{i=1}^{k+1} x_i^2 = R^2,x_{k+2}=0 \}$

Point $M$ outside of the plan: $ 0 \not = a_{k+2}$

Can we find $(b_1,...,b_{n+2})$ the center and $R'$ the radius of a new sphere containing both?

By definition that sphere needs to have: $\sum_{i=1}^{k+2} (x_i-b_i)^2 = R'^2$ and $\sum_{i=1}^{k+2} (a_i-b_i)^2 = R'^2$.

The first equation is equivalent to $\sum_{i=1}^{k+2} x_i^2- 2x_ib_i + b_i^2 =R^2 - 2\sum_{i=1}^{k+2}x_ib_i + \sum_{i=1}^{k+2}b_i^2 = R^2 + b_{k+2}^2 = R'^2$ after setting all the relevant $b_i$ to 0 apart from the last one.

The second equation becomes $\sum_{i=1}^{k+2} (a_i-b_i)^2 = \sum_{i=1}^{k+2} a_i^2 - 2a_{k+2}b_{k+2} + R'^2-R^2 = R'^2 \Leftrightarrow b_{k+2} = \frac{1}{2a_{k+2}}(R^2 - \sum_{i=1}^{k+2} a_i^2)$

That gives you your solutions, but you need to be careful about the different conditions hidden here and there ($R' > R$, etc.)

You'll notice it's very similar to the case in $\mathbb{R^3}$.

Matt B.
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