Free Equation of a Sphere Calculator

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A sphere's standard-form equation, (x-h)²+(y-k)²+(z-l)²=r², directly encodes its center (h,k,l) and radius r — the natural extension of a circle's equation into three dimensions. This is used in 3D modeling and CAD software when defining spherical primitives, in physics when describing the boundary of a field or particle's influence, and in computational geometry when checking whether a point lies inside, on, or outside a spherical region. Given the expanded "general form" instead (with all the squared terms multiplied out), completing the square in each variable recovers the center and radius.

How it works

Choose center+radius (to build the equation directly by substituting into the standard form) or general form (to recover center and radius). For general form, expand x²+y²+z²+Dx+Ey+Fz+G=0 and complete the square on each variable separately: the center coordinates are h=-D/2, k=-E/2, l=-F/2, and the radius is found from r²=h²+k²+l²-G.

Enter direction.
Enter center h.
Enter center k.
Enter center l.
Enter radius (r).
Enter d (coefficient of x).
Enter e (coefficient of y).
Enter f (coefficient of z).
Enter g (constant).
Click Calculate to see your results.

Examples

Center (1,2,3), radius 5

Substituting directly into the standard form gives the equation (x-1)²+(y-2)²+(z-3)²=25.

General form x²+y²+z²-4x+6y-10z+13=0

Completing the square: h=-(-4)/2=2, k=-6/2=-3, l=-(-10)/2=5. Then r²=2²+(-3)²+5²-13=4+9+25-13=25, so r=5. The sphere is centered at (2,-3,5) with radius 5.

Common mistakes to avoid

Forgetting to halve and negate the coefficients when completing the square (h = -D/2, not -D).
Applying the formula to a general-form equation whose x², y², z² coefficients aren't all 1 without first dividing through by that common coefficient.
Mixing up the sign of G in r²=h²+k²+l²-G — it is subtracted, not added.

Frequently asked questions

Those coefficients don't describe a real sphere — a negative "radius squared" has no real solution, so the calculator flags this case as invalid.

That describes a degenerate "sphere" that has collapsed to a single point — technically valid algebraically, but not a true sphere with any surface area.

A circle is the 2D version — it has a center (h,k) and radius but lives in a flat plane. A sphere adds a third coordinate (l) and exists in full 3D space.

Yes, and their coefficients must all be equal to 1 (or a common factor you've divided out first) — if x², y², and z² have different coefficients, the surface isn't a sphere but a more general ellipsoid.

Four non-coplanar points are enough to solve for a sphere's equation uniquely, analogous to how three non-collinear points determine a circle in 2D — though this particular calculator instead works directly from a known center and radius, or from a general-form equation.