General equation of a sphere is given by x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0, where ( − u, − v, − w) is the centre and 
is the radius of the sphere.
Let P be a any plane and F is the foot of perpendicular from centre(c) of the sphere to this plane.
If CF > 
then plane P neither touches nor cuts the sphere.
If CF = 
then plane P touches the sphere.
If CF < 
then intersection of plane P and sphere is a circle with
radius = 
(i) Find the equation of the sphere having centre at (1, 2, 3) and touching the plane
x + 2y + 3z = 0.
Text Solution
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(i) Given plane is x + 2y + 3z = 0 ..... (1)
Let H be the centre of the required sphere.
Given H ≡ (1, 2, 3)
Radius of the sphere,
HP = length of perpendicular from H to plane (1)
= 
= 
Equation of the required sphere is (x – 1) 2 + (y – 2) 2 + (z – 3) 2 = 14
or x 2 + y 2 + z 2 – 2x – 4y – 6z = 0
(ii) OP ⊥ AP
α ( α – 1) + β ( β – 2) + γ ( γ – 3) = 0
∴ Locus of P( α , β , γ ) is
∴ P ( α , β , γ )
x 2 + y 2 + z 2 – x – 2y – 3z = 0
(iii) The given line can be written as 
= λ ⇒ x = λ , y = 2 λ , z = 3 λ
subtitute these values in sphere
λ 2 + 4 λ 2 + 9 λ 2 – 2 λ – 4 λ – 
= 0 ⇒ λ = 0, 2
so the points are (0, 0, 0) and (2, 4, 6) so length of chord = 
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