A sphere that contains a circle and a point 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?
 A: 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 $Q$.
A: 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).
A: 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.
