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.
In Standard Form
The standard equation of a sphere given the center and radius is:
(x – h)2 + (y – k)2 + (z – l)2 = r2, here (x, y, z) = any point on the sphere, (h, k, l) = center, r = radius
Equation of a Sphere
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 = r2
⇒ x2 + y2 + z2 = r2
Let us derive the equation.
Derivation
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 = r2 (applying the Pythagorean theorem)
So the equation of sphere with center and radius is:
(x – h)2 + (y – k)2 + (z – l)2 = r2
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:
x2 + y2 + z2 = r2
∴r = √( x2 + y2 + z2)
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 = r2, here (h, k, l) = (1, 1, 2), r = 5 units ∴ The equation is: (x – 1)2 + (y – 1)2 + (z – 2)2 = 52
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 = r2, 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 = 112.
Solution:
Given the equation of the sphere: x² + (y – 1)² + (z – 4)² = 11² ∴ Center (h, k, l) = (0, 1, 4), Given r2 = 11² ⇒ Radius (r) = √(11²) ∴ r = 11 units
In General Form
The equation of a sphere in general (expanded) form is written as:
x2 + y2 + z2 + 2ux + 2vy + 2wz + 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 = u2 + v2 + w2 – r2
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: x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, here (-u, -v, -w) = (1, 2, 3) ∴ u = -1, v = -2, w = -3 Now, d = u2 + v2 + w2 – r2, 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
