Sphere:

A sphere is a locus of a point which moves in the space such that the distance from a fixed point is constant. Fixed point is called as centre of sphere and constant distance is called as radius of sphere.

Equation of Sphere in Different Forms:

• If centre of sphere is (a, b, c) and radius is r, then the equation of sphere is

      (x - a)² + (y - b)² + (z - c)² = r².

• If centre of sphere is the origin o and radius is r, then x² + y² + z² = r².
• General form: The general equation of sphere can be given as x² + y² + z² + 2ux + 2vy + 2wz + d = 0

Centre of sphere = (-u, -v, -w), radius = .

• Diameter form: The equation of 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.

Example. 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                                                                     ... (1)

            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) => 2 u - 6v + d = - 10            ... (4)

            solving these 4 equations we get,

            u = - 1, v = 3, w = - 2 and d = 10

            Therefore required equation of sphere is

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

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

Solution :     Radius of sphere = 

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

                        That is. x² + y² + z² - 4x + 6y - 8z - 22 = 0.

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