Analytical Solid Gepmetry

Chapter 3

Analytical Solid Geometry

3.5   Spheres

fig 3.5-1
The distance between center (x1, y1, z1) and any point (x, y, z) of a sphere is radius r.
√ (x - x1)2 + (y - y1)2 + (z - z1)2   = r


Therefore the equation of the sphere with center (x1, y1, z1 and radius r is
(x - x1)2 + (y - y1)2 + (z - z1)2 = r2

Example 11.
Find the equation of the plane tangent to the sphere
(x - 2)2 + (y - 1)2 + (z + 1)2 = 14
at the point (3, 4, 1).

Solution
fig 3.5-2

Directed values of the line joining center C(2, 1, -1) of the sphere and the given point P(3, 4, 1) are
CP = (1, 3, 2).
Line CP is perpendicular to the tangent plane. Thus the equation of the plane is
x + 3y + 2z = d.
Since P(3, 4, 1) is on this plane, so we get
3 + 3(-1) + 2(1) = d
d = 17
The equation of the plane is
x + 3y + 2z = 17.
Example 12.
Find the equation of the sphere with center (0, 1, 0) and touching the plane x - 2y + 2z + 5 = 0.

Solution
The equation of the line that passes through the center C(0, 1, 0) and perpendicular to the plane x - 2y + 2z + 5 = 0 is
x - 0 ⁄ 1 = y - 1 ⁄ -2 = z - 0 ⁄ 2
x ⁄ 1 = y - 1 ⁄ -2 = z ⁄ 2
So coordinates of the points on the line are of the form
x = s
y = 1 + (-2)s = 1 - 2s
z = 2s
If one of these points P is on the plane, then
s - 2(1 - 2s) + 2(2s) + 5 = 0
s   -   2 + 4s   +   4s   +   5 = 0
            9s = -3
            s = - 1⁄3
Therefore the point of intersection is P(- 1⁄3, 5⁄3, -2⁄3). Thus we have the radius
CP = √ (- 1⁄3 - 0)2 + (5⁄3 - 1)2 + (- 2⁄3 - 0)2
      = √ 1⁄9 + 4⁄9 + 4⁄9
      = 1

Therefore the equation of the sphere is (x2 + (y - 1)2 + z2 = 1.
Example 13.
Find the equation of a sphere that passes through the points (9, 0, 0), (3, 13, 5) and (11, 0, 10), given that its center lies on the yz-plane.
solution
The equation of the sphere with center (x1, y1, z1) and radius r is
(x - x1)2 + (y - y1)2 + (z - z1 )2 = r2
Distance from center (x1, y1, z1) to each of the given point is
(9 - 0)2 + (0 - y1)2 + (0 - z1)2 = r2     (in yz-plane, x1 = 0)
81 + y12 + z12 = r2 (1)
(3 - 0)2 + (13 - y1)2 + (5 - z1)2 = r2
9 + (13 - y1)2 + (5 - z1)2 = r2 (2)
(11 - 0)2 + (0 - y1)2 + (10 - z1)2 = r2
121 + y12 + (10 - z1)2 = r2 (3)
From (1) and (3)
81 + y12 + z12   =   121 + y12 + (10 - z1)2
81 + y12 + z12   =   121 + y12 + 100 - 20z1 + z12
0   =   40 + 100 - 20z1
z1 = 7

From (1) and (2),
81 + y12 + z12   =   9 + (13 - y1)2 + (5 - z1)2
81 + y12 + 72   =   9 + 169 - 26y1 + y12 + (5 - 7)2
130   =   182 - 26y1
y1   =   2
Therefore x1 = 0, y1 = 2 and z1 = 7 and substitute these values in equation (1), we get
r2 = 134.

The equation of the sphere is
(x - 0)2 + (y - 2)2 + (z - 7)2 = 134.

Exercise 3.5
  1. Find the equation of sphere with center C and radius r.
    2. (a)  C(1, -2, 4),   r = 3
    (b)   C(2, 6, -3),   r = 2
    (c)   C(2, 3, 5), r = 5

  2. Check whether the given point P lies inside, outside or on a sphere.
    3. (a)   Center C(0, 0, 0), radius r = 3 and point P(1, 1, 1)
    (b)   Center C(0, 0, 0), radius r = 3 and point P(2, 1, 2)
    (c)   Center C(0, 0, 0), radius r = 3 and point P(10, 10, 10)

  3. Find the equation of the sphere on the join of (1, -1, 1) and (-3, 4, 5) as diameter.

  4. Find the equation of the plane tangent to the sphere
    (x + 2)2 + (y - 1)2 + (z + 3)2 = 27
    at the point (3, 2, -2).

  5. Find the equation of the sphere with center (6, -7, -3) and touching the plane
    4x - 2y - z = 17.

  6. What is the equation of the sphere which passes through the points (3, 0, 2), (-1, 1, 1) and (2, -5, 4) and whose center lies on the plane 2x + 3y + 4z = 6 ?

Complex Numbers
Exercise 3.5
Answers
1.   Find the equation of sphere with center C and radius r.
(a)  C(1, -2, 4),   r = 3
Solution
x1 = 1,   y1 = -2,   z1 = 4
(x - x1)2 + (y - y1)2 + (z - z1)2 = r2
(x - 1)2 + (y - (-2))2 + (z - 4)2 = 32
(x - 1)2 + (y + 2)2 + (z - 4)2 = 9.

(b)   C(2, 6, -3),   r = 2
x1 = 2,   y1 = 6,   z1 = -3
(x - x1)2 + (y - y1)2 + (z - z1)2 = r2
(x - 2)2 + (y - 6)2 + (z - (-3))2 = 22
(x - 2)2 + (y - 6)2 + (z + 3)2 = 4.

(c)   C(2, 3, 5), r = 5
x1 = 2,   y1 = 3,   z1 = 5
(x - x1)2 + (y - y1)2 + (z - z1)2 = r2
(x - 2)2 + (y - 3)2 + (z - 5)2 = 52
(x - 2)2 + (y - 3)2 + (z - 5)2 = 25

2.   Check whether the given point P lies inside, outside or on a sphere.
(a)   Center C(0, 0, 0), radius r = 3 and point P(1, 1, 1)
Solution
Note: Calculate the distance between the point and the sphere's center. If the distance is smaller than the radius (r) of the sphere, the point is inside the plane.
If the distance is equal to the radius (r) of the sphere, the point is on the sphere.
And if the distance is greater than the radius (r) of the sphere, the point is outside the sphere.

x1 = 0,   y1 = 0,   z1 = 0, and
x = 1,   y = 1,   z = 1
distance CP = √ (x - x1)2 + (y - y1)2 + (z - z1)2  
= √ (1 - 0)2 + (1 - 0)2 + (1 - 0)2
= √ 12 + 12 + 12
= √ 3 < (r = 3)
The given point P lies inside the sphere.

(b)   Center C(0, 0, 0), radius r = 3 and point P(2, 1, 2)
distance CP = √ (x - x1)2 + (y - y1)2 + (z - z1)2  
= √ (2 - 0)2 + (1 - 0)2 + (2 - 0)2
= √ 22 + 12 + 22
= √ 9
= 3
= (r = 3)
The given point P lies on the sphere.

(c)   Center C(0, 0, 0), radius r = 3 and point P(10, 10, 10)

distance CP = √ (x - x1)2 + (y - y1)2 + (z - z1)2  
= √ (10 - 0)2 + (10 - 0)2 + (10 - 0)2
= √ 102 + 102 + 102
= √ 300
= 3 √ 3   > (r = 3)

The given point P lies outside the sphere.

3.   Find the equation of the sphere on the join of (1, -1, 1) and (-3, 4, 5) as diameter.
Solution
diameter = √ (x2 - x)2 + (y2 - y)2 + (z2 - z)2
= √ (-3 - 1)2 + (4 - (-1))2 + (5 - 1)2
= √ (-4)2 + 52 + 42
= √ 16 + 25 + 16
= √ 57
radius = diameter⁄2
= √ 57 &fracl; 2
center = ( 1 + (-3)⁄2 , -1 + 4⁄2 , 1 + 5⁄2 )
= ( -2⁄2 , 3⁄2 , 6⁄2 )
= (-1, 3⁄2, 3)
The equation of the sphere with center (-1, 3⁄2, 3) and radius √ 57 &fracl; 2 is
(x - 1)2 + (y - 3⁄2)2 + (z - 3)2 = ( √ 57 &fracl; 2)2
= 57 &fracl; 4
4.   Find the equation of the plane tangent to the sphere
(x + 2)2 + (y - 1)2 + (z + 3)2 = 27
at the point (3, 2, -2).
Solution
Equation of the sphere with center (x1, y1, z1) and radius r is
(x - x1)2 + (y - y1)2 + (z - z1)2 = r2
(x + 2)2 + (y - 1)2 + (z + 3)2 = 27
- x1 = + 2 ,   - y1 = - 1 ,   - z1 = + 3   and   r2 = 27
∴ x1 = - 2 ,   y1 = 1 ,   z1 = - 3   and   r = √ 27
center = (x1, y1, z1)
= (-2, 1, -3)
      r = √ 27
Directed values of the line joining center C(-2, 1, -3) of the sphere and the given point P(3, 2, -2) are
CP = (x2 - x1,   y2 - y1 ,   z2 - z1)
= (3 - (-2),   2 - 1,   -2 - (-3) )
= (5, 1, 1)
Line CP is perpendicular to the tangent plane.
Thus equation of the plane is ax + by + cz = d.
5x + 1y + 1z = d
5x + y + z = d
Since P(3, 2, -2) is on this plane, so we get
5(3) + 2 - 2 = d
15 = d
d = 15
The equation of the plane is 5x + y + z = 15.
5.   Find the equation of the sphere with center (6, -7, -3) and touching the plane
4x - 2y - z = 17.
Solution
The equation of the line that passes through the center (6, -7, -3) and perpendicular to the plane 4x - 2y - z = 17 is
x - 6⁄4 = y - (-7)⁄-2 = z - (-3)⁄-1
x - 6⁄4 = y + 7⁄-2 = z + 3⁄-1 = s
x - 6⁄4 = s ,     y + 7⁄-2 = s ,     z + 3⁄-1 = s
x - 6 = 4s , y + 7 = -2s , z + 3 = -s
∴ x = 6 + 4s , y = -7 - 2s , z = -3 - s
If one of these points P is on the plane, then
4x - 2y - z = 17.
4(6 + 4s)   -   2(-7 - 2s)  -   (-3 - s) = 17
24 + 16s   +   14 + 4s   +   3 + s = 17
41 + 21s = 17
21s = -24
s = -8⁄7
x = 6 + 4s   =   6 + 4 (-8⁄7)   =   42 - 32⁄7   =   10⁄7
y = -7 - 2s   =   -7 - 2 (-8⁄7)   =   -49 + 16⁄7   =   -33⁄7
z = -3 - s   =   -3 - (-8⁄7)   =   -21 + 8⁄7   =   -13⁄7
∴ P ( 10⁄7 , -33⁄7 , -13⁄7 )
radius = √ (10⁄7 - 6)2 + (-33⁄7 + 7)2 + (- 13⁄7 + 3)2
= √ (10 - 42⁄ 7 )2 + (-33 + 49⁄7 )2 + (-13 + 21⁄7 )2
= √ (-32⁄ 7 )2 + (16⁄7 )2 + (8⁄7 )2
= √ 1024⁄ 49 + 256⁄49 + 64⁄49
= √ 1344⁄ 49
The equation of the sphere is
(x - 6)2 + (y - (-7))2 + (z - (-3))2 = ( √ 1344⁄ 49   )2
(x - 6)2 + (y + 7)2 + (z + 3)2 = 1344⁄ 49
= 192⁄ 7

6.   What is the equation of the sphere which passes through the points (3, 0, 2), (-1, 1, 1) and (2, -5, 4) and whose center lies on the plane 2x + 3y + 4z = 6 ?
Solution
Center of the sphere C(x1, y1, z1) lies on the plane 2x + 3y + 4z = 6.
2x1 + 3y1 + 4z1 = 6     ______(1)

Distance from center (x1, y1, z1) to the given point (3, 0, 2) is
(3 - x1)2 + (0 - y1)2 + (2 - z1)2 = r2
9 - 6x1 + x12   +   y12   +   4 - 4z1 + z12   = r2
13 - 6x1 + x12   +   y12   -   4z1 + z12   = r2     ______(2)

Distance from center (x1, y1, z1) to the given point (-1, 1, 1) is
(-1 - x1)2 + (1 - y1)2 + (1 - z1)2 = r2
1 + 2x1 + x12   +   1 - 2y1 + y12   +   1 - 2z1 + z12   = r2
3 + 2x1 + x12   -   2y1 + y12   -   2z1 + z12   = r2     ______(3)

Distance from center (x1, y1, z1) to the given point (2, -5, 4) is
(2 - x1)2 + (-5 - y1)2 + (4 - z1)2 = r2
4 - 4x1 + x12   +   25 + 10y1 + y12   +   16 - 8z1 + z12   = r2
45 - 4x1 + x12   +   10y1 + y12   -   8z1 + z12   = r2     ______(4)

From (2) and (3)
13 - 6x1 + x12   +   y12   -   4z1 + z12   = 3 + 2x1 + x12   -   2y1 + y12   -   2z1 + z12
13 - 6x1   -   4z1   = 3 + 2x1   -   2y1   - 2z1
-6x1 - 2x1   + 2y1   - 4z1 + 2z1   = 3 - 13
-8x1 + 2y1 - 2z1 = -10
2 (-4x1 + y1 - z1) = -10
-4x1 + y1 - z1 = -5     ______(5)

From (2) and (4)
13 - 6x1 + x12   +   y12   -   4z1 + z12   = 45 - 4x1 + x12   +   10y1 + y12   -   8z1 + z12
13 - 6x1   -   4z1   = 45 - 4x1   + 10y1   -   8z1
-6x1 + 4x1   -10y1   - 4z1 + 8z1   = 45 - 3
-2x1 - 10y1 + 4z1   = 32
2 (-x1 - 5y1 + 2z1)   = 32
-x1 - 5y1 + 2z1   = 16     ______(6)

From (3) and (4)
3 + 2x1 + x12   -   2y1 + y12   -   2z1 + z12   = 45 - 4x1 + x12   +   10y1 + y12   -   8z1 + z12
3 + 2x1   -   2y1   -   2z1   = 45 - 4x1   +   10y1   -   8z1
2x1 + 4x1   - 2y1 - 10y1   -2z1 + 8z1   = 54 - 3
6x1 - 12y1 + 6z1   = 42
6 (x1 - 2y1 + z1)   = 42
x1 - 2y1 + z1   = 7     ______(7)

(6) + (7)
-x1 - 5y1 + 2z1   = 16
  x1 - 2y1 +   z1   =   7
        -7y1 + 3z1   = 23     ______(8)

(6)×2   +   (1)
-2x1 - 10y1 + 4z1 = 32
  2x1 + 3y1 + 4z1 =   6
          -7y1 + 8z1 = 38     ______(9)

(9) - (8)
-7y1 + 8z1 = 38
-7y1 + 3z1 = 23
            5z1 = 15
              z1 = 3
Substituting z1 = 3 in ______(8)
-7y1 + 3z1   = 23
-7y1 + 3(3) = 23
-7y1 = 23 - 9
y1 = 14⁄-7
y1 = -2

Substituting y1 = -2   and z1 = 3 in ______(7)
x1 - 2y1 + z1   = 7     ______(7)
x1 - 2(-2) + 3 = 7
x1 + 4 + 3 = 7
x1 + 7 = 7
x1 = 0

center (x1, y1, z1) = (0, -2, 3)
(x - x1)2 + (y - y1)2 + (z - z1)2 = r2
(3 - 0)2 + (0 - (-2))2 + (2 - 3)2 = r2
9 + 4 + 1 = r2
r2 = 14

The equation of the sphere is
(x - 0)2 + (y - (-2))2 + (z - 3)2 = 14
x2 + (y + 2)2 + (z - 3)2 = 14