A ball’s circular slice depends on how close the plane passes to its center. Can we find its radius without measuring the slice? For our radius-13 cm model and a plane 5 cm from the center, a 5–12–13 triangle supplies the answer. First distinguish the surface from the solid.
Sphere: boundary; ball: interior too
Prerequisites: circles and disks, Pythagoras, a normal to a plane, spatial coordinates.
A sphere centered at C with radius R>0 consists of points exactly R from C. The closed ball consists of points at distance at most R, including the boundary. The center belongs to the ball, not the sphere. This extends the distinction between a circle and its filled disk into space.
Rotating a filled half-disk about its diameter creates a ball; rotating the semicircular arc creates a sphere. A real hollow shell has thickness and material between two surfaces; a mathematical sphere has no thickness. A diameter passes through the center and joins sphere points, with length 2R.
Why the section is a circle
Let plane α be distance d from C, and H the perpendicular foot from C onto α. For every X in the plane, CH⊥HX, so (CX)²=d²+(HX)². X lies on the sphere precisely when CX=R, giving:
(HX)^(2) = (R)^(2) − (d)^(2).
If d<R, these points form a circle centered at H with radius ρ=√(R²−d²). Both directions are proved: every common point belongs to the circle, and every circle point is R from C. For the ball, CX≤R instead gives the filled disk with that boundary.
Circle on a sphere: R=13, d=5, ρ=12
Section CHX: two legs and a radius
Intersection, tangency, no intersection
In our example R=13 cm and d=5 cm, so ρ=√(169−25)=12 cm. The section circumference is 2πρ=24π cm; the filled disk’s area is πρ²=144π≈452.39 cm². This is a planar section’s area, not the whole sphere’s area or a patch of its curved surface.
At d=0, H=C and ρ=R: a plane through the center gives a great circle of the sphere and great disk of the ball. It has the largest radius among planar circular sections. As d increases, ρ decreases; one perspective photograph cannot measure this relationship.
A tangent plane: one common point
The formula immediately gives the other cases. At d>R the right-hand side is negative, so there is no intersection. At d=R, HX=0 leaves only H. This is a tangent plane to the sphere, and H the tangency point.
The tangent plane is perpendicular to the radius at that point. Conversely, a plane through sphere point T perpendicular to CT meets the sphere only at T. For any other point X in that plane, (CX)²=(CT)²+(TX)²>R². Thus every sphere point has one tangent plane, specified by point T and normal CT.
Tangency is not merely touching the outline in a screen image. The spatial criterion is distance R from the center or perpendicularity to the corresponding radius.
Coordinates check the construction
For our center C(1;−2;3), R=13, the sphere equation is:
(x − 1)^(2) + (y + 2)^(2) + (z − 3)^(2) = 169.
Plane z=8 is |8−3|=5 from the center. Substitution gives (x−1)²+(y+2)²=144, together with z=8: a circle centered at H(1;−2;8). Without the z constraint, the equation in x and y would describe an infinite cylindrical surface.
Plane z=16 is distance 13 from C and tangent at T(1;−2;16). Plane z=19 is distance 16 and misses the sphere. Replacing = by ≤ in the sphere equation gives the ball, not another sphere. One coordinate frame thus checks definitions and all three intersection cases.
Curved surface areas and volumes require other formulas, coming in G034. Here the task is to identify a circle, filled disk, surface or solid before calculating.
Sources and limits
Sphere versus ball, the three cases and tangent perpendicularity: Merzliak and coauthors, 2019, section 21, pp.133–135. Additionally, OpenStax, Calculus Volume 3, §2.2 defines a sphere by equal distances, derives its equation, and explains why a missing z constraint gives a cylinder. The 13/5/12 model, coordinates and Pythagorean proof are our own.