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Coordinates and Quadric Surfaces · Tutorial 4 of 999

Spheres from Center-Radius Data

Translate center-radius information into a sphere equation, read its geometry, and verify points by checking their distance from the center.

Beginner 9 min read

What You'll Learn

  • Write the center-radius equation of a sphere in three-dimensional coordinates.
  • Identify a sphere’s center and radius directly from its equation.
  • Find a sphere’s radius when given its center and a point on the sphere.
  • Test whether a specified point lies on a sphere.
  • Find points on a sphere when some coordinates are fixed.
  • Distinguish the sphere’s surface from the solid ball it bounds.

From a Center and Radius to an Equation

A sphere in space is the three-dimensional analogue of a circle in a plane. Its defining information is a center and a radius: every point on the sphere is exactly the same distance from the center. Coordinates let us turn that geometric description into an equation.

In this tutorial, a point in space is written as an ordered triple. Let the center be \(C=(h,k,l)\), and let a variable point be \(P=(x,y,z)\). The three-dimensional distance formula says that the distance from \(C\) to \(P\) is \(\sqrt{(x-h)^2+(y-k)^2+(z-l)^2}\). A point belongs to the sphere precisely when this distance equals its radius.

Definition: The sphere with center \(C=(h,k,l)\) and radius \(r>0\) is the set of all points \(P=(x,y,z)\) whose distance from \(C\) is \(r\). Its center-radius equation is
$$ (x-h)^2+(y-k)^2+(z-l)^2=r^2. $$

The left side is the square of the distance from \((x,y,z)\) to \((h,k,l)\). The equation therefore says that the distance is \(r\). Notice that the radius is squared in the equation, so the right side is \(r^2\), not \(r\).

Why the Equation Describes Exactly the Sphere

The formula is more than a convenient pattern: it is equivalent to the geometric definition. The distance formula gives the distance from an arbitrary point to the proposed center. Equating that distance to \(r\) and squaring both sides produces the center-radius equation. Squaring is reversible here because distances and \(r\) are nonnegative.

1
Start with the distance condition.
A point \(P=(x,y,z)\) is on the sphere exactly when its distance from \(C=(h,k,l)\) is \(r\).
2
Use the distance formula.
Write the condition as \(\sqrt{(x-h)^2+(y-k)^2+(z-l)^2}=r\).
3
Square both sides.
Since both sides are nonnegative, the condition is equivalent to \((x-h)^2+(y-k)^2+(z-l)^2=r^2\).

This also proves the converse. If a point satisfies the equation, then its squared distance from the center is \(r^2\). Since distance and \(r\) are nonnegative, its distance is \(r\), so the point lies on the sphere. Thus the equation includes every point on the sphere and excludes every point not on it.

Reading the Center and Radius

In center-radius form, compare each squared coordinate difference with the pattern \((x-h)^2+(y-k)^2+(z-l)^2=r^2\). The constants subtracted from \(x\), \(y\), and \(z\) give the center coordinates. The positive square root of the right side gives the radius.

Key takeaway: In \((x-h)^2+(y-k)^2+(z-l)^2=r^2\), the center is \((h,k,l)\), and the radius is \(r\). The sign inside each parenthesis matters: \(x+3\) can be written \(x-(-3)\), so it indicates a center coordinate of \(-3\).

The radius is a length, so it is taken to be positive for a sphere. The equation with \(r=0\) describes only the center point, rather than a sphere of positive radius. A negative right side cannot equal a sum of squares, so it describes no real points.

Worked Example: Write an Equation from a Given Center and Radius

Find the equation of the sphere with center \(C=(2,-1,3)\) and radius \(4\). Then verify that \(P=(6,-1,3)\) lies on it.

Substitute \(h=2\), \(k=-1\), \(l=3\), and \(r=4\) into the center-radius form:

$$ (x-2)^2+(y-(-1))^2+(z-3)^2=4^2. $$

Simplify the signs and square the radius:

$$ (x-2)^2+(y+1)^2+(z-3)^2=16. $$

To check \(P=(6,-1,3)\), substitute its coordinates into the left side:

$$ (6-2)^2+(-1+1)^2+(3-3)^2 =4^2+0^2+0^2 =16. $$

The result equals the right side, \(16\), so \(P\) lies on the sphere. Equivalently, its distance from the center is \(\sqrt{16}=4\), the specified radius.

When the Center and a Point Are Given

Sometimes the radius is not supplied directly. If the center and any one point on the sphere are known, the radius is the distance between them. After finding that distance, place its square on the right side of the center-radius equation. The given point should then satisfy the resulting equation, providing a useful check.

Worked Example: Find the Radius from a Point on the Sphere

A sphere has center \(C=(-1,2,0)\) and passes through \(Q=(2,6,0)\). Find its equation and verify the given point.

First calculate the distance from \(C\) to \(Q\):

$$ r=\sqrt{(2-(-1))^2+(6-2)^2+(0-0)^2} =\sqrt{3^2+4^2+0^2} =\sqrt{9+16} =\sqrt{25} =5. $$

The center is \((-1,2,0)\), and the radius is \(5\). Substitution into the center-radius form gives

$$ (x-(-1))^2+(y-2)^2+(z-0)^2=5^2, $$ $$ (x+1)^2+(y-2)^2+z^2=25. $$

Check the given point by substitution:

$$ (2+1)^2+(6-2)^2+0^2 =3^2+4^2 =9+16 =25. $$

The point satisfies the equation, as required. Its distance from the center is \(5\), so it is consistent with the calculated radius.

Testing Whether a Point Lies on a Sphere

To test a proposed point, substitute its three coordinates into the left side of the equation. If the result equals the right side, the point lies on the sphere. If the two values differ, it does not. This is exactly a comparison of squared distances: equality means the point is at the sphere’s radius from the center.

Worked Example: Find Points with Two Coordinates Fixed

Find the points on the sphere centered at \((1,2,-2)\) with radius \(3\) that have \(x=1\) and \(y=2\).

The equation of the sphere is

$$ (x-1)^2+(y-2)^2+(z+2)^2=9. $$

Set \(x=1\) and \(y=2\). The first two squared terms are both zero, leaving

$$ (1-1)^2+(2-2)^2+(z+2)^2=9, $$ $$ (z+2)^2=9. $$

Thus \(z+2=3\) or \(z+2=-3\), so \(z=1\) or \(z=-5\). The two points are \((1,2,1)\) and \((1,2,-5)\).

Verify each point in the sphere equation. For \((1,2,1)\), the left side is \(0^2+0^2+3^2=9\). For \((1,2,-5)\), it is \(0^2+0^2+(-3)^2=9\). Both equal the right side, so both points lie on the sphere.

Common Sign and Radius Mistakes

A frequent error is to read the signs in the parentheses as the center coordinates without accounting for subtraction. For example, a learner might claim that \((x+3)^2+(y-2)^2+(z+1)^2=25\) has center \((3,2,1)\). That is incorrect: \(x+3=x-(-3)\), and \(z+1=z-(-1)\). The correct center is \((-3,2,-1)\), and the radius is \(\sqrt{25}=5\).

Another common error is to use \(25\) as the radius because it appears on the right side. It is the squared radius. The radius is \(5\). For a direct check of the center interpretation, the point \((-3,2,4)\) differs from \((-3,2,-1)\) by \(0\), \(0\), and \(5\), so its distance from the center is \(\sqrt{0^2+0^2+5^2}=5\). Substituting it into the equation gives \(0^2+0^2+5^2=25\), confirming that it lies on the sphere.

Sphere Surface and Solid Ball

The equation with equality describes the sphere’s surface: points exactly \(r\) units from the center. It does not describe the interior. The solid ball consists of points whose distance from the center is at most \(r\), so its coordinate condition uses an inequality instead:

$$ (x-h)^2+(y-k)^2+(z-l)^2\le r^2. $$

The distinction is useful when interpreting a problem. “On the sphere” means equality; “inside or on the sphere” means the inequality. In either case, the center and squared distance organize the coordinate calculation.

Check Your Understanding

Use the center-radius form and the distance formula to answer these questions.

  1. Write the equation of a sphere centered at \((0,-2,4)\) with radius \(3\).
  2. What are the center and radius of \((x-5)^2+(y+1)^2+(z-2)^2=36\)?
  3. A sphere has center \((1,0,-2)\) and passes through \((1,3,-2)\). What is its equation?
  4. Does \((2,1,0)\) lie on the sphere \((x-2)^2+(y-1)^2+z^2=9\)? Show the substitution.
  5. What is the difference between the equation of a sphere’s surface and the condition describing its solid ball?