Intersection of Plane with Sphere is Circle

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Theorem

The intersection of a plane with a sphere is a circle.


Proof

Let $S$ be a sphere of radius $R$ whose center is located for convenience at the origin.

Let $P$ be a plane which intersects $S$ but is not a tangent plane to $S$.

It is to be shown that $S \cap P$ is a circle.


Let $S$ and $P$ be embedded in a (real) cartesian space of $3$ dimensions.

Let this space be rotated until $P$ is parallel to the plane $z = 0$.

Thus from the Equation of Plane we have that $P$ can be described as:

$z = c$

where $\size c < R$ (or $P$ would not intersect $S$).

Let $A = \tuple {x, y, z}$ be an arbitrary point on $S \cap P$.

We have that:

\(\displaystyle x^2 + y^2 + z^2\) \(=\) \(\displaystyle R^2\) Equation of Sphere
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2 + y^2 + c^2\) \(=\) \(\displaystyle R^2\) Equation of Plane: $z - c$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2 + y^2\) \(=\) \(\displaystyle R^2 - c^2\)
\(\displaystyle \) \(>\) \(\displaystyle 0\) as $-R < c < R$

The result follows from Equation of Circle.

$\blacksquare$


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