Question

Find the equation of the sphere through the circle x^2+y^2+z^2 =9 , 2x+2y+2z =5 and the point (1, 2, 3).​

Answer 1

$\large\underline{\bold{Solution-}}$

$\tt \: Let \: S_1 : \: {x}^{2} + {y}^{2} + {z}^{2} - 9 = 0 - - - (1)$

and

$\tt \: Let \: P_1 : \: 2x + 2y + 2z - 5 = 0 - - (2)$

Now,

The required equation of sphere which passes through the intersection of circle (1) and (2) is

$\bf \: S_1 \: + \: k \:P_1 \: = \: 0$

$\tt \: {x}^{2} + {y}^{2} + {z}^{2} - 9 \: + \: k \: (\: 2x + 2y + 2z - 5) = 0 - - (3)$

Now,

It is given that (3) passes through (1, 2, 3), so

$\tt \: {(1)}^{2} + {(2)}^{2} + {(3)}^{2} - 9 \: + \: k \: (\: 2 \times 1 + 2 \times 2 + 2 \times 3 - 5) = 0$

$\tt \: 1 + 4 + 9 + k(2 + 4 + 6 - 5) = 0$

$\tt \: 14 + 7k = 0$

$\tt \: 7k = - 14$

$\therefore \: \tt \: k \: = \: - \: 2$

So,.

required equation is

$\tt \: {x}^{2} + {y}^{2} + {z}^{2} - 9 - 2 \: (\: 2x + 2y + 2z - 5) = 0$

$\tt \: {x}^{2} + {y}^{2} + {z}^{2} - 9 - 4x - 4y - 4z + 10 = 0$

$\tt \: {x}^{2} + {y}^{2} + {z}^{2} - 4x - 4y - 4z + 1 = 0$

Additional Information :-

1. Properties of a sphere

• A sphere is perfectly symmetrical

• It is not a polyhedron

• All the points on the surface are equidistant from the centre.

• It does not have a surface of centres

• It has constant mean curvature

• It has a constant width and circumference.

2. Let us consider a sphere having centre (a, b, c) and radius 'r', then equation of sphere is

$\tt \: {(x - a)}^{2} + {(y - b)}^{2} + {(z - c)}^{2} = {r}^{2}$

3. The general equation of sphere is given by

$\tt \: {x}^{2} + {y}^{2} + {z}^{2} + 2ux + 2vy + 2wz + c = 0 \: having$

$\tt \: Centre \: = \: (- \: u, \: - \: v, \: - \: w)$

and

$\tt \: radius \: (r) \: = \: \sqrt{ {u}^{2} + {v}^{2} + {w}^{2} - c }$