Relative Position of Two Spheres: Tangent and Intersecting

The relative position of two spheres is decided by comparing the distance d between their centers with the radii r₁ and r₂: they are separate if d > r₁ + r₂, externally tangent if d = r₁ + r₂, intersect in a circle if |r₁ − r₂| < d < r₁ + r₂, internally tangent if d = |r₁ − r₂|, and one lies inside the other if d < |r₁ − r₂|. In brief: the spheres x² + y² + z² = 25 and (x − 6)² + y² + z² = 25 have d = 6 < 10, so they meet in a circle in the plane x = 3, with the radius 4.

This guide gives the five cases, the radical plane (the plane of the common circle or the common tangent plane), the center and radius of the intersection circle, and the tangency point, with worked examples and a table.

What are the possible positions of two spheres?

Let the centers be C₁ and C₂ with d = |C₁C₂|, and assume r₁ ≥ r₂. The cases are the same as for two circles in the plane, because the whole configuration is symmetric about the line C₁C₂.

• d > r₁ + r₂: separate, no common points; the gap is d − r₁ − r₂.
• d = r₁ + r₂: externally tangent at one point.
• r₁ − r₂ < d < r₁ + r₂: intersecting in a circle.
• d = r₁ − r₂ > 0: internally tangent at one point.
• d < r₁ − r₂: one sphere inside the other, no common points (concentric if d = 0).

Example 1: classifying pairs

The pairs of spheres centered at the origin with r₁ and at (d, 0, 0) with r₂:

• r₁ = 1, r₂ = 2, d = 5: 5 > 3, separate (the gap is 2).
• r₁ = 3, r₂ = 2, d = 5: 5 = 5, externally tangent.
• r₁ = 5, r₂ = 5, d = 6: 0 < 6 < 10, intersecting.
• r₁ = 3, r₂ = 2, d = 1: 1 = 3 − 2, internally tangent.
• r₁ = 5, r₂ = 2, d = 1: 1 < 3, the small sphere is inside.

Example 2: from general equations

S₁: x² + y² + z² − 4x + 2y − 4 = 0 has the center (2, −1, 0) and r₁ = 3. S₂: x² + y² + z² + 4x − 4y + 4 = 0 has the center (−2, 2, 0) and r₂² = 4 + 4 + 0 − 4 = 4, so r₂ = 2. The distance d = √(16 + 9) = 5 = r₁ + r₂, so the spheres are externally tangent.

What is the radical plane?

Subtracting the equations of two spheres (in the general form with the same leading coefficients) cancels the quadratic terms and leaves a linear equation: the radical plane. It is perpendicular to the line of centers and consists of the points with equal power with respect to both spheres. For intersecting spheres, it contains the common circle; for tangent spheres, it is the common tangent plane at the point of tangency.

Example 3: the radical plane in Example 2

S₁ − S₂: (−4x + 2y − 4) − (4x − 4y + 4) = −8x + 6y − 8 = 0, that is 4x − 3y + 4 = 0. Its normal (4, −3, 0) is parallel to the line of centers (−4, 3, 0) ✓.

How do you find the point of tangency?

For externally tangent spheres, the point lies on the line of centers at the distance r₁ from C₁: T = C₁ + (r₁/d)(C₂ − C₁). For internally tangent spheres, the same formula holds when r₁ is the bigger radius.

Example 4: the point of tangency in Example 2

T = (2, −1, 0) + (3/5)(−4, 3, 0) = (−2/5, 4/5, 0). Check on S₂: (−2/5 + 2)² + (4/5 − 2)² = 64/25 + 36/25 = 4 ✓, and on the radical plane: 4(−2/5) − 3(4/5) + 4 = −8/5 − 12/5 + 4 = 0 ✓.

Example 5: an internal tangency

Spheres x² + y² + z² = 9 and (x − 1)² + y² + z² = 4: d = 1 = 3 − 2. T = (0, 0, 0) + (3/1)(1, 0, 0) = (3, 0, 0). Check on the second sphere: (3 − 1)² = 4 ✓.

How do you describe the intersection circle?

The common circle lies in the radical plane, and its center is on the line of centers. If the center of the circle is at the distance x₁ from C₁ towards C₂, then x₁ = (d² + r₁² − r₂²)/(2d), and the radius is ρ = √(r₁² − x₁²). The derivation uses the Pythagorean theorem for the two right triangles with the common leg ρ.

Example 6: two equal spheres

x² + y² + z² = 25 and (x − 6)² + y² + z² = 25. Subtracting: 12x − 36 = 0, so the radical plane is x = 3. Then x₁ = (36 + 25 − 25)/12 = 3 and ρ = √(25 − 9) = 4. The circle: x = 3, y² + z² = 16.

Example 7: two different spheres

x² + y² + z² = 25 and (x − 5)² + y² + z² = 16. d = 5, x₁ = (25 + 25 − 16)/10 = 3.4, ρ = √(25 − 11.56) = √13.44 ≈ 3.67. Via the radical plane: subtract the equations: 10x − 25 = 25 − 16, giving x = 3.4 ✓.

Example 8: the common volume’s cross-section

For Example 6, the intersection circle has the area πρ² = 16π ≈ 50.3. The lens-shaped region between the spheres has the circle as its widest cross-section.

When are two spheres orthogonal?

Two intersecting spheres are orthogonal (their tangent planes at the common points are perpendicular) if and only if d² = r₁² + r₂², because the radii to a common point are then perpendicular.

Example 9: an orthogonal pair

The spheres with centers (0, 0, 0) and (5, 0, 0) and radii 3 and 4: d² = 25 = 9 + 16. The spheres are orthogonal, and the 3-4-5 triangle explains it. The common circle has x₁ = (25 + 9 − 16)/10 = 1.8 and ρ = √(9 − 3.24) = 2.4, the height of the 3-4-5 triangle on its hypotenuse.

Table of cases

Where does the radical plane come from?

The radical axis of two circles and the radical plane of two spheres were studied in the early nineteenth century in the French school of geometry; the term “radical axis” is attributed to Louis Gaultier (1813), and the concept of power of a point is associated with Jakob Steiner. Constructions with spheres, such as inversion, led to the study of orthogonal spheres.

Which mistakes are common?

Mixing up the tangent cases

External tangency needs d = r₁ + r₂, internal tangency d = |r₁ − r₂|. Computing the difference of the wrong radii is the common slip.

Subtracting equations with different scaling

Both equations must have the same coefficient of x² + y² + z² (normally 1) before subtracting, otherwise quadratic terms survive.

Reading the radical plane as the circle

The radical plane contains the circle; to describe the circle, add one of the sphere equations.

Forgetting the nested case

When d < r₁ − r₂, the system has no real solution, even though the spheres “overlap” in volume.