Table of Contents

Last modified on August 3rd, 2023

A sphere is a 3 – dimensional object. It is the collection of all the points equidistant from a fixed point, the center. It can be referred to as the 3-d version of a circle. So its equation is almost similar to that of a circle with an added variable for the extra dimension.

The standard equation of a sphere given the center and radius is:

**(x – h) ^{2} + (y – k)^{2} + (z – l)^{2} = r^{2}**, here (x, y, z) = any point on the sphere, (h, k, l) = center, r = radius

When the center is at the origin (0, 0, 0), the equation of a sphere in standard form becomes:

(x – 0)^{2} + (y – 0)^{2} + (z – 0)^{2} = r^{2}

⇒ **x ^{2 }+ y^{2} + z^{2} = r^{2}**

Let us derive the equation.

Let (h, k, l) be a fixed point in the 3 – D plane.

Let (x, y, z) be another point from (h, k, l) at a distance r such that r is constant and a real positive number.

Squaring both sides,

(x – h)^{2} + (y – k)^{2} + (z – l)^{2} = r^{2} (applying the Pythagorean theorem)

So the equation of sphere with center and radius is:

**(x – h) ^{2} + (y – k)^{2} + (z – l)^{2} = r^{2}**

r = √[(x – h)^{2} + (y – k)^{2} + (z – l)^{2} ] ——— (applying the distance formula)

If the center is at the origin (0, 0, 0)

Then the equation of the sphere is:

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

∴r = √( x^{2 }+ y^{2} + z^{2})

Let us solve some examples to understand the above concept.

**Find the equation of sphere with center and radius as (1, 1, 2) and 5 units respectively.**

Solution:

As we know,

The equation of a sphere in standard form is:

(x – h)^{2} + (y – k)^{2} + (z – l)^{2} = r^{2}, here (h, k, l) = (1, 1, 2), r = 5 units

∴ The equation is:

(x – 1)^{2} + (y – 1)^{2} + (z – 2)^{2} = 5^{2}

**Write the equation of a sphere given the center as (2, 4, 6) and radius 3 units.**

Solution:

As we know,

The equation of a sphere in standard form is:

(x – h)^{2} + (y – k)^{2} + (z – l)^{2} = r^{2}, here (h, k, l) = (2, 4, 6), r = 3 units

∴ The equation is:

(x – 2)² + (y – 4)² + (z – 6)² = 3²

Finding the center and radius of a sphere when the **EQUATION **is known

**Find the center and radius of a sphere having the equation x² + (y – 1)² + (z – 4)² = 11 ^{2}.**

Solution:

Given the equation of the sphere: x² + (y – 1)² + (z – 4)² = 11²

∴ Center (h, k, l) = (0, 1, 4),

Given r^{2} = 11²

⇒ Radius (*r*) = √(11²)

∴ r = 11 units

The equation of a sphere in general (expanded) form is written as:

**x**^{2} + **y**^{2} + **z**^{2} + 2**ux**** + 2****vy**** + 2****wz**** + ****d**** = 0**, here (-u, -v, -w) = center, and u, v, w, & d = constants

**∵** Radius (*r*)** = ${\sqrt{u^{2}+v^{2}+w^{2}-d}}$,** here (-u, -v, -w) = center, and u, v, w, & d = constants, then

d = u^{2} + v^{2} + w^{2} – r^{2}

Let us solve some examples to understand the above concept.

**Write the equation of a sphere in expanded form centered at (1, 2, 3) with a radius of 6 units.**

Solution:

As we know,

The equation of a sphere in expanded form is:**x**^{2} + **y**^{2} + **z**^{2} + 2**ux**** + 2****vy**** + 2****wz**** + ****d**** = 0**, here (-u, -v, -w) = (1, 2, 3)

∴ u = -1, v = -2, w = -3

Now, d = u^{2} + v^{2} + w^{2} – r^{2}, here u = -1, v = -2, w = -3, r = 6 units

∴ d = 1 + 4 + 9 – 36

= -22

So, the equation is:

x² + y² + z² – 2x – 4y – 6z – 22

Last modified on August 3rd, 2023