# 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:

\(\ds x^2 + y^2 + z^2\) | \(=\) | \(\ds R^2\) | Equation of Sphere | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds x^2 + y^2 + c^2\) | \(=\) | \(\ds R^2\) | Equation of Plane: $z - c$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds x^2 + y^2\) | \(=\) | \(\ds R^2 - c^2\) | |||||||||||

\(\ds \) | \(>\) | \(\ds 0\) | as $-R < c < R$ |

The result follows from Equation of Circle.

$\blacksquare$

## Sources

- 1976: W.M. Smart:
*Textbook on Spherical Astronomy*(6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $2$.*The spherical triangle.*