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How to prove that a circle and a point outside the plane of the circle determine a sphere?

I know that the circle is determined by three non-collinear points, so from the circle, we get 3 non-collinear points and we also have an extra point which is outside the plane of the circle, therefore, we have 4 non-coplanar points. These will determine a sphere.

But how do we prove that the sphere that we formed contains all of the circle from before?

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    There are already answers, but to see why it works intuitively, put a sphere on your circle, and make its radius vary, and also its position, "over" or "under" the circle). What portion of space will be eventually "hit" by the sphere? Answer: all space except the plane where the circle lie. And it suggests a way to actually determine the sphere. – Jean-Claude Arbaut Sep 04 '14 at 10:57

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Here's a different construction that makes the inclusion of the circle in the sphere manifest.

Start with a circle $C$ and a point $Q$. First, the set of points which are equidistant from all points on $C$ is the line $L$ which crosses the plane of $C$ perpendicularly and passes through its center. Now pick a point $Q'$ on $C$, and draw the plane $P$ which perpendicularly bisects the line segment $QQ'$; all points on $P$ are equidistant from $Q$ and $Q'$. Then the line $L$ and plane $P$ intersect at some point $Q''$ so long as $L$ is not parallel to $P$ (note that this only occurs if $C$ and $Q$ lie in the same plane, i.e. no sphere.) But $Q''$ is equidistant from all points on $C$ along with the exterior point $Q$, and so we can construct a sphere centered at $Q''$ which includes both $C$ and $Q$.

Semiclassical
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  • Thank You! Can this construction be generalized to higher dimensions? I wish to ask if it possible to find a $k+1$-sphere containing a $k$-sphere and a point outside the $k$-sphere. –  Sep 04 '14 at 06:12
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    @Vivek there's only one line perpendicular to the $k$-sphere going through its center in the $k+1$-dimensional subspace containing the sphere and the point. A line intersecting with a `$k$-dimensional space in a $k+1$-dimensional space generally results in a pointlike intersection - the center of the $k+1$-dimensional sphere. – John Dvorak Sep 04 '14 at 06:16
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Note that three non-collinear points determine both a plane and a unique circle in that plane. Intersecting the plane with a sphere containing the the three points gives a circle (intersection of plane with sphere) containing those three points, which must be the same circle (because it was unique).

Mark Bennet
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    So is your argument generalizable for higher dimensional cases? For example: Given in $\mathbb{R}^n$, with a sufficiently large $n\in\mathbb{N}$, a $p$-sphere $\mathcal{C}$, with $p<n-1$ and a point $\mathbf{a}\in\mathbb{R}^n$ that lies outside the $(p+1)$-plane containing $\mathcal{C}$. Then, there always exists a $(p+1)$-sphere that contain both $\mathcal{C}$ and $\mathbf{a}$. Moreover, there always exist $k$-spheres, for each $k=p+1,\ldots,n-1$ that will contain $\mathcal{C}$ and $\mathbf{a}$. – Ayan Sep 04 '14 at 07:13
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I am curious about finding a geometric way to construction, but below is my algebraic approach.

We assume the function of sphere is $$(x-a)^2+(y-b)^2+(z-c)^2=R^2$$ with $$\begin{cases} (x_0,y_0,z_0)\\ x^2+y^2=r^2,z=0 \end{cases}$$ being its solution.

Then we have two facts $$(x_0-a)^2+(y_0-b)^2+(z_0-c)^2=R^2\cdots(1)$$ $$(x-a)^2+(y-b)^2+c^2=R^2 \Longleftrightarrow x^2+y^2=r^2\cdots(2)$$

From $(2)$, we have $a=0$, $b=0$ and $R^2-c^2=r^2$. Then we take these into $(1)$, and we can solve what $c$, $R$ are.

By using the algebraic method, we can get the conclusion that the sphere is existed and unique.

gaoxinge
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