Sphere

A sphere is a locus of a point which moves in space such that its distance from a fixed point is constant. Fixed point is called centre of sphere and constant distance is called radius of sphere. A sphere is the result of rotation of a circle about one of its parameters. Most of the properties and constituents of a sphere are similar to that of a circle.     

There is a very minute difference between the properties of a sphere and a circle. We begin with some of the components of a sphere:

Some Important Terminology

• A line segment joining two points on a sphere and passing through the centre is called a diameter.

• The shortest possible distance between any two points on a sphere is termed as geodesic.

• The surface area of the sphere of radius ‘r’ is given by A = 4πr²

• The volume of the sphere of radius ‘r’ is given by V = 4/3 πr³

• Great Circle:

?Great circle is an important concept associated with the sphere. The circle which is formed as a result of intersection of the surface of the sphere with a plane that passes through the center of the sphere. There can be infinite number of great circles satisfying a particular requirement.

Equation of Sphere in Diverse Forms

Now we discuss the equation of sphere in various diverse forms: 
The general equation of sphere with center at (a, b, c) and radius ‘r’ is given by

(x – a)² + (y – b)² + (z – c)² = r²

The same general equation in expanded form can be written as

x² + y² + z² + 2ux + 2vy + 2wz + d = 0

The center in this case becomes (–u, –v, –w),

And radius is given by r= √μ² + v² + w² – d. 

Remark:

• If instead of (a, b, c), the centre is at origin i.e. (0, 0, 0) then the equation becomes x² + y² + z² = r².

• In spherical coordinates, the points on the sphere can be written as 

x = x₀ + r cos θ sin φ

y = y₀ + r cos θ sin φ

z = z₀ + r cos φ where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π

Diameter form: 

By the diameter form we mean the equation of the sphere when extremities of the diameter are given. 

Equation of a sphere whose extremities of diameter are A (x₁, y₁, z₁) and B (x₂, y₂, z₂) is

(x – x₁) (x – x₂) + (y – y₁) (y – y₂) + (z – z₁) (z – z₂) = 0.

You may also view the following video for more on sphere

We now illustrate some of the examples based on these concepts: 

Illustration:

Find the equation of the sphere which passes through the points (1, –3, 4), (1, –5, 2) and (1, –3, 0) and whose centre is on the plane x + y + z = 0. 

Solution:

Let equation of the sphere be

x² + y² + z² + 2ux + 2vy + 2wz + d = 0 

Its centre is (- u, - v, - w) which is on x + y + z = 0

⇒ u + v + w = 0                                                                   

It passes through (1, - 3, 4) ⇒ 2u - 6v + 8w + d = - 26       … (2)

                            (1, - 5, 2) ⇒ 2 u - 10 v + 4 w + d = - 30   … (3)

and it passes through (1, - 3, 0) ⇒2u - 6v + d = - 10  … (4)

Solving these four equations we get,

u = - 1, v = 3, w = - 2 and d = 10

Therefore required equation of the sphere is

  x² + y² + z² - 2x + 6y - 4z + 10 = 0.  

Illustration:

Find the equation of the sphere whose centre is (2, –3, 4) and which passes through the point (1, 2, –1).

Solution:

We know that the radius of the sphere is

Radius of sphere = √{(2–1)² + (–3–2)² + (4+1)²} = √51

∴ Equation of the sphere is (x – 2)² + (y + 3)² + (z – 4)² = (√51)²

i.e. x² + y² + z² – 4x + 6y – 8z – 22 = 0.  

Students are advised to master these concepts in order to remain competitive in the IIT JEE and other engineering exams.

Related Resources

• Click here for the Entire Syllabus of IIT JEE Mathematics.

• Look into the Past Year Papers with Solutions to get a hint of the kinds of questions asked in the exam.

• You can get to know the Important Books of Mathematics here.

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