A sphere is a round 3D shape where every point on its surface is the same distance from a central point. This distance is called the radius, and the central point is the centre of the sphere. The diameter is the straight line that passes through the centre and touches two points on the surface—it's twice the radius. You can think of a sphere as what you get when you spin a circle around its diameter.

Just like a circle has a radius and diameter, so does a sphere—but in three dimensions. Since a sphere is a solid shape, it has both surface area (the area covering the outside) and volume (the space it takes up inside).

Parts of Sphere: We know a sphere is a three-dimensional object, which is round in shape. So before moving to the equation of sphere and other things related to the sphere. We will first learn about some important parts of the sphere which are as follows:

• Radius of sphere: It is the distance from the exact centre of the sphere to any point on the surface of the sphere.

• Diameter of sphere: It is the longest line segment that can be drawn between two points on the sphere. Its length is twice the radius of the sphere, i.e. diameter = 2\(r\).

• Circumference of sphere: It is defined as the length of the great circle of the sphere. A great circle is one that contains the diameter of the sphere. It is the largest possible circle that can be drawn inside a sphere.

• Surface area of the sphere: It is the region covered by a surface of a spherical object in a three-dimensional space.

• Volume of the sphere: It is the capacity of a sphere. Or in other words, it is the amount of air that a sphere can be held inside it.

What is the Equation of Sphere?

A sphere is a 3D shape made up of all points that are at the same distance (called the radius) from a fixed point known as the centre. If the centre of the sphere is at the point (a, b, c) and the radius is ‘r’, the equation of the sphere is:

  (x - a)² + (y - b)² + (z - c)² = r²

This is called the general form of the sphere’s equation.

We can also write this equation in an expanded form:

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

In this form, the centre of the sphere is (-u, -v, -w), and the radius is:

  r = √(u² + v² + w² - d)

Note: If the centre is at the origin (0, 0, 0), the equation becomes:

  x² + y² + z² = r²

This is known as the standard form of the equation of a sphere.

Equation of sphere in diametric form

Diametric form of sphere means that the equation of the sphere when extremities of the diameter are given. So, equation of sphere whose extremities of diameter are \(A(x_{1},y_{1},z_{1})\) and \(B(x_{2},y_{2},z_{2})\) is given as

\((x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})+(z-z_{1})(z-z_{2})=0\).

Equation of sphere in cylindrical coordinates

In cylindrical coordinates, we have

X = r cos(p), Y = r sin(p) and Z = z,

Here, ‘r’ is the radius in the x-y plane and ‘p’ is the angle in the x-y plane.

We know that the equation of the sphere in the cartesian coordinates system is \(x^{2}+y^{2}+z^{2}=r^{2}\), where ‘r’ is the radius of the sphere.

Since \(x^{2}+y^{2}=r^{2}\) in cylindrical coordinates, the equation of sphere in cylindrical coordinates can be written as

\(r^{2}+z^{2}=R^{2}\),

Equation of sphere in the parametric form

The common form of the parametric equation of a sphere is:

\((x,y,z) = (acos\theta sin\phi, asin\theta sin\phi, acos\phi)\),

where, a is the constant radius, \(\theta\in [0,2\pi)\) is the longitude and \(\phi\in[0,\pi]\) is the co-altitude.

Since the surface of a sphere is two-dimensional, parametric equations usually will have two variables (i.e. \(\theta\) and \(\phi\)).

Equation of sphere in vector form

The general vector equation of sphere with centre C having position vector ‘c’ and radius is ‘a’

is given as

\((\textbf{r}-\textbf{c})^{2}=a^{2}\),

i.e., \(\textbf{r}^{2}-2\textbf{r}\cdot \textbf{c}+(\textbf{c}^{2}-\textbf{a}^{2})=0\).

Circumference of Sphere

The circumference of a sphere is the circumference of the largest possible circle inside the sphere (a great circle). It is also known as the great circle which contains the centre and the diameter of the sphere. It is measured in linear units such as mm, cm, m, in, or ft. The below figure shows the circumference of a sphere.

The formula to find the circumference of a sphere is given by

Circumference (C) = 2\(\pi\)r, where \(\pi\) = 22/7 = 3.141 and r = radius.

Surface Area of Sphere

Surface area of sphere is the area covered by the outer surface of the sphere. We know that the sphere is a completely curved shape, therefore the curved surface area is equal to the total area of the sphere (i.e surface area). It is also known as the lateral surface area of a sphere.

It is measured in square units. The below figure shows the surface area of a sphere.

Formula for the surface area of a sphere is given by

Surface area of a sphere = 4\(\pi r^{2}\), where r is the radius of the sphere.

If the diameter of a sphere is given, the formula for the surface area of a sphere is given by

Surface area of a sphere = \(\pi d^{2}\), where d is the diameter of the sphere.

Learn about Surface Area of Cuboid

The volume of a sphere is the region occupied inside the sphere. It can simply be defined as the capacity of the sphere. The volume of a sphere depends on the radius of a sphere and it gets increased by increasing the radius and vice-versa. It is measured in cubic units. The below figure shows the volume of a sphere.

Formula for volume of a sphere is given by

Volume of a sphere (V) = \(\frac{4}{3}\pi r^{3}\), where \(\pi\) = 22/7 = 3.141 and r = radius.

Equation of Sphere Passing Through Four Points

Let the equation of sphere be

\(x^{2}+y^{2}+z^{2}+2ux+2vy+2wz+d=0\) …………..(1)

Let this sphere pass through four points \((x_{1},y_{1},z_{1})\), \((x_{2},y_{2},z_{2})\), \((x_{3},y_{3},z_{3})\), and \((x_{4},y_{4},z_{4})\). Thus,

\(x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+2ux_{1}+2vy_{1}+2wz_{1}+d=0\), ………..(2)

\(x_{2}^{2}+y_{2}^{2}+z_{2}^{2}+2ux_{2}+2vy_{2}+2wz_{2}+d=0\), ………..(3)

\(x_{3}^{2}+y_{3}^{2}+z_{3}^{2}+2ux_{3}+2vy_{3}+2wz_{3}+d=0\), ………..(4)

and

\(x_{4}^{2}+y_{4}^{2}+z_{4}^{2}+2ux_{4}+2vy_{4}+2wz_{4}+d=0\). ………..(5)

Eliminate u, v, w and d from equations (1), (2), (3), (4) and (5), and we get

\(\begin{vmatrix}x^{2}+y^{2}+z^{2}&2x &2y &2z &1 \\x_{1}^{2}+y_{1}^{2}+z_{1}^{2}&2x_{1} &2y_{1} &2z_{1} &1 \\x_{2}^{2}+y_{2}^{2}+z_{2}^{2}&2x_{2} &2y_{2} &2z_{2} &1 \\x_{3}^{2}+y_{3}^{2}+z_{3}^{2}&2x_{3} &2y_{3} &2z_{3} &1 \\x_{4}^{2}+y_{4}^{2}+z_{4}^{2}&2x_{4} &2y_{4} &2z_{4} &1 \\\end{vmatrix}\) = 0.

On solving the above determinant, we get the equation of sphere passing through four points (i.e. \((x_{1},y_{1},z_{1})\), \((x_{2},y_{2},z_{2})\), \((x_{3},y_{3},z_{3})\), and \((x_{4},y_{4},z_{4})\)).

Difference Between Circle and Sphere

Difference between a circle and sphere are given below:

Properties of a Sphere

• A sphere looks the same from every direction; it is perfectly round and fully symmetrical.

• It has a smooth, curved surface with no flat parts.

• A sphere does not have any corners (vertices) or straight edges.

• Every point on the outside of the sphere is at the same distance from the center. This fixed distance is called the radius.

• A sphere is not a polyhedron because polyhedrons have flat faces, edges, and corners, which a sphere does not.

• Air bubbles form a spherical shape because a sphere uses the least amount of surface area to hold a certain volume of air.

• Among all shapes with the same surface area, a sphere can hold the most volume.

• The formula to calculate the volume of a sphere is:

  Volume = (4/3) × π × r³ cubic units,
  where r is the radius of the sphere.

Solved Examples of Equation of Sphere

Example 1: Find the volume of the sphere whose radius is 3 cm.

Solution: We know that the volume of a sphere is

V = \(\frac{4}{3}\pi r^{3}\)

V = \(\frac{4}{3}\pi 3^{3}\)

V = \(4\times \pi \times 3^{2}\)

V = 113.076 \(cm^{3}\).

Example 2: Find the surface area of the sphere whose radius is 5 units.

Solution: We know that the surface area of a sphere is

Surface area of a sphere = 4\(\pi r^{2}\)

Surface area of a sphere = 4\(\pi 5^{2}\)

Surface area of a sphere = 314.1 \(units^{2}\).

Example 3: Find the equation of the sphere passing through the points (0,0,0), (-1,2,0), (0,1,-1) and (1,2,5).

Solution: Let the equation of the sphere be

\(x^{2}+y^{2}+z^{2}+2gx+2fy+2hz+c=0\)

Since it passes through (0,0,0), (-1,2,0), (0,1,-1) and (1,2,5), we have

c = 0, -2g + 4f = -5, 2f – 2h = -2, 2g + 4f + 10h = -30.

On solving these equations, we get

f = -5/2, g = -5/2, and h = -3/2.

Hence, the equation of the required sphere is

\(x^{2}+y^{2}+z^{2}-5x-5y-3z=0\).

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