In short

A sphere is the locus of all points at a fixed distance (the radius) from a fixed point (the centre) in three-dimensional space. Its standard equation is (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2, where (a, b, c) is the centre and r is the radius. Expanding this gives the general form x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0, from which you can read off the centre as (-u, -v, -w) and the radius as \sqrt{u^2 + v^2 + w^2 - d}. When a plane intersects a sphere, the cross-section is always a circle.

Hold a cricket ball at arm's length. Every point on its surface is the same distance from the centre — about 36 mm. That distance is the radius. The surface itself — the leather, the seam, everything you can see and touch — is the sphere.

In mathematics, a "sphere" refers to the surface only, not the solid ball inside. The solid is called a ball or a solid sphere. This distinction matters: the equation of a sphere describes the surface, and a point is "on the sphere" only if it is exactly r units from the centre — not closer, not further.

A circle is the set of all points in a plane at a fixed distance from a centre. A sphere is the exact same idea lifted into three dimensions. If you already understand the equation of a circle — (x - h)^2 + (y - k)^2 = r^2 — then the equation of a sphere will look almost identical, with one extra squared term for the third coordinate.

The equation of a sphere

Start from the definition. A sphere has centre C = (a, b, c) and radius r > 0. A point P = (x, y, z) lies on the sphere if and only if

|CP| = r

The distance from C to P is \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}. Setting this equal to r and squaring both sides:

Standard equation of a sphere

A sphere with centre (a, b, c) and radius r has the equation

(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2

A point (x_0, y_0, z_0) lies on the sphere if and only if it satisfies this equation.

This is the standard form. You can read the centre and radius directly: the centre is the point that appears inside the parentheses (with signs flipped), and the radius is the square root of the right-hand side.

For example, (x - 3)^2 + (y + 1)^2 + (z - 5)^2 = 49 is a sphere with centre (3, -1, 5) and radius 7.

The simplest case: a sphere centred at the origin with radius r has the equation

x^2 + y^2 + z^2 = r^2

The unit sphere — centre at origin, radius 1 — is x^2 + y^2 + z^2 = 1.

The general form

Expand the standard form. Take (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2:

(x^2 - 2ax + a^2) + (y^2 - 2by + b^2) + (z^2 - 2cz + c^2) = r^2
x^2 + y^2 + z^2 - 2ax - 2by - 2cz + (a^2 + b^2 + c^2 - r^2) = 0

Write u = -a, v = -b, w = -c, and d = a^2 + b^2 + c^2 - r^2. Then:

x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0

This is the general form of the equation of a sphere. It arises naturally when a problem gives you an expanded equation and asks you to identify the sphere.

Going the other way — from general to standard form — is completing the square, three times over:

Centre: (-u, -v, -w)

Radius: r = \sqrt{u^2 + v^2 + w^2 - d}

For this to represent a real sphere, the quantity under the square root must be positive: u^2 + v^2 + w^2 - d > 0. If it equals zero, the "sphere" degenerates to a single point (the centre). If it is negative, no real points satisfy the equation.

A sphere with centre C and radius r A perspective drawing of a sphere. The centre C is marked with a dot. A radius line extends from C to a point P on the surface. A dashed circle shows the equator of the sphere. The coordinate axes are drawn to indicate the 3D setting. C(a,b,c) P(x,y,z) r x z y
A sphere with centre $C(a, b, c)$ and radius $r$. Every point $P$ on the surface satisfies $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$. The dashed ellipses show cross-sections — every such cross-section is a circle.

Finding centre and radius from the general form

The most common task in exams: you are given an equation and asked to identify the sphere.

The recipe is always the same. Group the x-terms, the y-terms, and the z-terms, complete the square in each group, and read off the centre and radius.

Take x^2 + y^2 + z^2 - 6x + 4y - 2z - 11 = 0. Compare with x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0:

2u = -6 \;\Rightarrow\; u = -3, \quad 2v = 4 \;\Rightarrow\; v = 2, \quad 2w = -2 \;\Rightarrow\; w = -1, \quad d = -11

Centre: (-u, -v, -w) = (3, -2, 1).

Radius: r = \sqrt{u^2 + v^2 + w^2 - d} = \sqrt{9 + 4 + 1 + 11} = \sqrt{25} = 5.

Or, completing the square directly:

(x^2 - 6x + 9) + (y^2 + 4y + 4) + (z^2 - 2z + 1) = 11 + 9 + 4 + 1
(x - 3)^2 + (y + 2)^2 + (z - 1)^2 = 25

Same answer: centre (3, -2, 1), radius 5. Both methods are equivalent — use whichever feels faster.

The diameter form

Sometimes you know the endpoints of a diameter and want the sphere's equation directly. If A = (x_1, y_1, z_1) and B = (x_2, y_2, z_2) are the endpoints of a diameter, then a point P = (x, y, z) lies on the sphere if and only if \vec{PA} \cdot \vec{PB} = 0.

Why? Because the angle in a semicircle is 90°. If AB is a diameter, then any point P on the sphere (other than A and B themselves) sees A and B at a right angle — the vectors from P to A and from P to B are perpendicular. A dot product of zero captures perpendicularity.

Write this out in coordinates:

\vec{PA} \cdot \vec{PB} = (x_1 - x)(x_2 - x) + (y_1 - y)(y_2 - y) + (z_1 - z)(z_2 - z) = 0

Expanding:

(x - x_1)(x - x_2) + (y - y_1)(y - y_2) + (z - z_1)(z - z_2) = 0

Diameter form of a sphere

If A(x_1, y_1, z_1) and B(x_2, y_2, z_2) are the endpoints of a diameter, the sphere's equation is

(x - x_1)(x - x_2) + (y - y_1)(y - y_2) + (z - z_1)(z - z_2) = 0

The centre is the midpoint of AB: \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right).

The radius is half the length of AB: r = \frac{1}{2}\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}.

This form is elegant because it bypasses the need to find the centre and radius first. You plug in the two endpoints and get the equation directly.

Diameter form: angle in a semicircle is 90 degrees A sphere shown in cross-section as a circle, with diameter AB. A point P on the sphere is connected to both A and B. The angle APB is 90 degrees, illustrating the theorem that the angle in a semicircle is a right angle. 90° A B P centre diameter
The angle-in-a-semicircle theorem: if $AB$ is a diameter, any point $P$ on the sphere sees $A$ and $B$ at a right angle. This is why $\vec{PA} \cdot \vec{PB} = 0$ captures exactly the points on the sphere.

Verification. Check that A and B lie on the sphere. Substituting A = (x_1, y_1, z_1): every factor becomes (x_1 - x_1) = 0 or (y_1 - y_1) = 0, so the entire left side is zero. The same happens for B. Both endpoints satisfy the equation, as expected.

Intersection of a plane and a sphere

This is one of the most important results about spheres: when a plane cuts through a sphere, the cross-section is always a circle.

A plane intersecting a sphere produces a circle A sphere is cut by a horizontal plane. The cross-section is a circle. The perpendicular distance from the sphere's centre to the plane is marked as d. The radius of the sphere is R, and the radius of the circular cross-section is r. A right triangle shows the relationship R squared equals r squared plus d squared. cutting plane C d r R M
A plane cuts the sphere at perpendicular distance $d$ from the centre $C$. The cross-section is a circle with radius $r = \sqrt{R^2 - d^2}$. The right triangle $CMr$ (with $R$ as hypotenuse) gives the relationship.

Proof. Let the sphere have centre C and radius R. Let the plane cut through the sphere, and let M be the foot of the perpendicular from C to the plane. The distance CM = d satisfies d < R (otherwise the plane does not cut through the sphere).

Take any point P on the intersection curve — the set of points that lie on both the sphere and the plane. Since P is on the sphere, CP = R. Since CM is perpendicular to the plane and P is in the plane, the triangle CMP is right-angled at M. By the Pythagorean theorem:

MP^2 = CP^2 - CM^2 = R^2 - d^2

The key observation: R^2 - d^2 is a constant. It does not depend on which point P you pick on the intersection. So every point P on the intersection is at the same distance \sqrt{R^2 - d^2} from M — and that is the definition of a circle with centre M and radius r = \sqrt{R^2 - d^2}.

Plane-sphere intersection

If a plane is at perpendicular distance d from the centre of a sphere of radius R, with d < R, the intersection is a circle of radius

r = \sqrt{R^2 - d^2}

The centre of this circle is the foot of the perpendicular from the sphere's centre to the plane.

When d = R, the plane is tangent to the sphere and the "circle" degenerates to a single point of tangency. When d > R, there is no intersection.

The great circle. When d = 0 — the plane passes through the centre of the sphere — the cross-section has the largest possible radius: r = R. This is called a great circle. The equator of the Earth is a great circle. Every geodesic (shortest path) on a sphere is an arc of a great circle, which is why flights between Delhi and San Francisco follow a curved path over the Arctic rather than a straight line on a flat map.

Three cross-sections of a sphere at different distances from the centre A sphere shown with three horizontal cross-sections. The middle one passes through the centre and produces the largest circle — the great circle with radius R. The upper and lower cross-sections produce smaller circles. The distances d1 and d2 from the centre to the upper and lower planes are shown. d₁ d₂ great circle (r = R) smaller circle smaller circle
Cross-sections of a sphere at different heights. The great circle (solid red) passes through the centre and has radius $R$. The dashed circles above and below are smaller: their radii are $\sqrt{R^2 - d^2}$, where $d$ is the distance from the centre to the cutting plane.

Finding the equation of the intersection circle

In coordinate geometry, you sometimes need to find the explicit equation of the circle of intersection. Here is the procedure.

Given: Sphere S: x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 and plane \Pi: lx + my + nz + p = 0.

Step 1. Find the centre C = (-u, -v, -w) and radius R = \sqrt{u^2 + v^2 + w^2 - d} of the sphere.

Step 2. Find the perpendicular distance from C to the plane:

d_\perp = \frac{|l(-u) + m(-v) + n(-w) + p|}{\sqrt{l^2 + m^2 + n^2}}

Step 3. The radius of the circle of intersection is r = \sqrt{R^2 - d_\perp^2}.

Step 4. The centre of the circle is the foot of the perpendicular from C to \Pi. If you need its coordinates, use the parametric form of the perpendicular line from C to \Pi and find where it meets \Pi.

The intersection circle lies simultaneously on the sphere and on the plane. In a coordinate system, you can describe it as the pair of simultaneous equations S = 0 and \Pi = 0.

Example 1: Find the centre and radius of the sphere x² + y² + z² − 4x + 6y − 8z + 4 = 0

Step 1. Identify the coefficients by comparing with x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0.

2u = -4 \;\Rightarrow\; u = -2, \quad 2v = 6 \;\Rightarrow\; v = 3, \quad 2w = -8 \;\Rightarrow\; w = -4, \quad d = 4

Why: the coefficient of x in the general form is 2u, so halving each linear coefficient gives u, v, w directly.

Step 2. Find the centre.

\text{Centre} = (-u, -v, -w) = (2, -3, 4)

Why: flipping the signs of u, v, w recovers the centre coordinates that were absorbed during expansion.

Step 3. Find the radius.

r = \sqrt{u^2 + v^2 + w^2 - d} = \sqrt{4 + 9 + 16 - 4} = \sqrt{25} = 5

Why: the radius formula comes from completing the square in all three variables simultaneously.

Step 4. Cross-check by completing the square explicitly.

(x^2 - 4x + 4) + (y^2 + 6y + 9) + (z^2 - 8z + 16) = -4 + 4 + 9 + 16
(x - 2)^2 + (y + 3)^2 + (z - 4)^2 = 25

Centre (2, -3, 4), radius 5. Matches.

Why: the completing-the-square method is equivalent to the formula method. Using both catches arithmetic errors.

Result: Centre = (2, -3, 4), Radius = 5.

The $z = 4$ cross-section of the sphere: a great circle with centre $(2, -3)$ and radius $5$. The dashed inner circle shows the cross-section at $z = 0$, with smaller radius $\sqrt{25 - 16} = 3$. Different slicing heights produce circles of different radii.

The two circles confirm the sphere's structure: the slice through the centre (z = 4) gives the largest circle (the great circle), while the slice at z = 0 — four units away from the centre — gives a smaller circle of radius \sqrt{25 - 16} = 3.

Example 2: Find the equation of the sphere with diameter endpoints A(1, 2, 3) and B(5, 6, −1), then find the circle of intersection with the plane x + y + z = 9

Step 1. Write the diameter form.

(x - 1)(x - 5) + (y - 2)(y - 6) + (z - 3)(z + 1) = 0

Why: the diameter form says \vec{PA} \cdot \vec{PB} = 0 — any point on the sphere sees the diameter at a right angle.

Step 2. Expand and simplify.

(x^2 - 6x + 5) + (y^2 - 8y + 12) + (z^2 - 2z - 3) = 0
x^2 + y^2 + z^2 - 6x - 8y - 2z + 14 = 0

Why: expanding each product and collecting like terms gives the general form of the sphere.

Step 3. Find the centre and radius.

Centre: midpoint of AB = \left(\frac{1+5}{2}, \frac{2+6}{2}, \frac{3-1}{2}\right) = (3, 4, 1).

Radius: r = \frac{1}{2}|AB| = \frac{1}{2}\sqrt{(5-1)^2 + (6-2)^2 + (-1-3)^2} = \frac{1}{2}\sqrt{16 + 16 + 16} = \frac{1}{2}\sqrt{48} = 2\sqrt{3}.

Check with the formula: u = -3, v = -4, w = -1, d = 14, so r = \sqrt{9 + 16 + 1 - 14} = \sqrt{12} = 2\sqrt{3}. Matches.

Why: the midpoint-and-distance method and the coefficient method both give the same centre and radius — a good cross-check.

Step 4. Find the circle of intersection with x + y + z = 9.

Perpendicular distance from centre (3, 4, 1) to the plane x + y + z - 9 = 0:

d_\perp = \frac{|3 + 4 + 1 - 9|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|-1|}{\sqrt{3}} = \frac{1}{\sqrt{3}}

Radius of the circle:

r_{\text{circle}} = \sqrt{R^2 - d_\perp^2} = \sqrt{12 - \frac{1}{3}} = \sqrt{\frac{35}{3}} = \frac{\sqrt{105}}{3} \approx 3.42

Why: the plane is close to the centre (distance \frac{1}{\sqrt{3}} \approx 0.577), so the cross-section is nearly a great circle — its radius is close to the sphere's radius 2\sqrt{3} \approx 3.46.

Result: The sphere is x^2 + y^2 + z^2 - 6x - 8y - 2z + 14 = 0, with centre (3, 4, 1) and radius 2\sqrt{3}. Its intersection with x + y + z = 9 is a circle of radius \frac{\sqrt{105}}{3}.

Projection onto the $xy$-plane. The solid circle is the great circle of the sphere (the $z = 1$ slice). The dashed red circle is approximately the intersection circle (whose radius is nearly equal to the sphere's radius because the cutting plane passes close to the centre). The dotted line is the trace of the cutting plane $x + y + z = 9$ in the $xy$-plane at $z = 1$.

The projection shows why the intersection circle is nearly as large as the sphere itself — the cutting plane barely misses the centre. If the plane passed exactly through (3, 4, 1), the intersection would be a great circle with radius 2\sqrt{3}. Moving the plane \frac{1}{\sqrt{3}} units away shaves the radius from 2\sqrt{3} to \frac{\sqrt{105}}{3} — a tiny reduction.

Common confusions

Going deeper

If you came here to learn the equations and the intersection formula, you have everything you need. The rest is for readers who want the tangent plane, the condition for two spheres to intersect, and the radical plane.

Equation of the tangent plane

At a point P_0 = (x_0, y_0, z_0) on the sphere (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2, the tangent plane is perpendicular to the radius CP_0. The direction of the radius is (x_0 - a, y_0 - b, z_0 - c), so the tangent plane is:

(x_0 - a)(x - a) + (y_0 - b)(y - b) + (z_0 - c)(z - c) = r^2

For the sphere in general form x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0, the tangent plane at (x_0, y_0, z_0) is obtained by the replacement rule — the same rule used for conics:

xx_0 + yy_0 + zz_0 + u(x + x_0) + v(y + y_0) + w(z + z_0) + d = 0

This is called the T = 0 form, using the same notation you may know from circles and conics.

Intersection of two spheres

If two spheres S_1 = 0 and S_2 = 0 intersect, what is their intersection curve?

Subtract one equation from the other: S_1 - S_2 = 0. Since the x^2, y^2, z^2 terms cancel, the result is a linear equation in x, y, z — which means it is a plane. This plane is called the radical plane of the two spheres.

The intersection of the two spheres is the intersection of this radical plane with either sphere — which, by the result proved above, is a circle. So two intersecting spheres meet in a circle, and the plane of that circle is S_1 - S_2 = 0.

This is a powerful idea. It means you can find the circle of intersection of two spheres without solving a quadratic — just subtract and intersect.

The power of a point

The power of a point P with respect to a sphere S is defined as the value S(P) — what you get when you substitute P's coordinates into the sphere's equation. If P is outside the sphere, the power is positive. If inside, negative. If on the sphere, zero.

For two spheres S_1 and S_2, the radical plane is the set of points whose power with respect to S_1 equals their power with respect to S_2: the equation S_1(P) = S_2(P), which is S_1 - S_2 = 0. This is the same plane found above.

The concept of power of a point is also used extensively for circles in 2D. The 3D version is a direct generalisation.

Spheres in Indian mathematics

Aryabhata, in the Aryabhatiya (499 CE), gave the formula for the volume of a sphere as \frac{4}{3}\pi r^3 — expressed in his own notation as the product of the area of a great circle and its own diameter, divided by a factor. While the equation-of-a-sphere formalism in Cartesian coordinates came much later with Descartes, the geometric understanding of the sphere as a shape generated by rotating a circle about a diameter was well established in Indian astronomical and mathematical traditions, where spherical geometry was essential for computing planetary positions and eclipses.

Where this leads next