spheres, tangent planes in 3-space

SPHERES AND TANGENT PLANES IN 3-SPACE

Notes and examples

Topics Covered in this Lesson

1) Equations of Spheres in 3-Space

2) Equations of Tangent Planes to Spheres in 3-Space

Equations of Spheres in 3-Space

A sphere is just a 3-dimensional circle and so is defined like the circle
by a center point and a radius, since the sphere is the locus of all points
that are exactly radius distance from the center point.

Standard Form of the Equation of a Sphere

Given the center ( x_o , y_o , z_o ) and the radius = r we write the equation
of the sphere as a statement about the square of the radius since we don't want to
deal with that big old ugly square root in the distance formula.

The equation of the sphere then is:

( x - x_o ) ² + ( y - y_o ) ² + ( z - z_o ) ² = r ²

which states that the square of the distance from any point
on the sphere to the center point equals the square of the radius.

example 1: Write the equation of the sphere with center ( -1, 3, 6) and radius = 5.

solution: ( x + 1 ) ² + ( y - 3) ² + ( z - 6 ) ² = 25.

example 2: Write the equation of the sphere with center ( - 4, 3, 7) through the origin.

solution: first we must find the square of the radius which is the distance
between the center point and the origin.

r² = 16 + 9 + 49 = 74 using the distance formula (which is really Pythagoras).
so the equation of the sphere is ( x + 4 ) ² + ( y - 3) ² + ( z - 7 ) ² = 74.

General Form of the Equation of a Sphere

To get the general form, we simply expand the standard form statement to get

x² + y² + z² - 2 x_ox - 2 y_oy - 2 z_oz + [(x_o)² + (y_o)² + (z_o)² - r² ] = 0.

Now, we collect up the constants (x_o)² + (y_o)² + (z_o)² - r² and set them = d
The coefficients of x, y and z which are -2(the center coordinates) get named a, b and c.
The equation now becomes

x² + y² + z² +ax + by + cz + d = 0

So, the coordinates of the center from the general form, are ( -½a, -½b, -½c )
and once we have the center, we find the radius using d = (x_o)² + (y_o)² + (z_o)² - r²

Note: the coefficients of x², y² and z² must = 1 for this to work. If they are not = 1,
divide the equation through by what they are, then use this approach.

example 3: Find the center and radius of the sphere x² + y² + z² +2x - 8y + 12z + 69 = 0

solution: the center is ( -½a, -½b, -½c ) = ( -1, 4, -6 )
Since d = 69 = (-1)² + (4)² + (-6)² - r² we solve for r and get r = 4.
The Standard form of this sphere's equation is ( x + 1 ) ² + ( y - 4) ² + ( z + 6 ) ² = 16

Had the equation in example 3 been presented as

3x² + 3y² + 3z² +6x - 24y + 36z + 217 = 0
we would divide the equation through by 3.

Equation of a Tangent Plane to a Sphere Given Point of Tangency

Since the tangent plane is perpendicular to the sphere's radius to the point of tangency, the radius vector serves as the normal for the tangent plane. Once we know the point of contact and the coordinates of the sphere's center, we have the normal vector and a point on the plane so we can find it's equation.

example 4: Write the equation of the tangent plane to the sphere
with equation ( x + 1 ) ² + ( y - 4) ² + ( z + 6 ) ² = 16 at the point P (- 4, 4, - 10).

solution: The center C is (-1, 4, -6 ). The radius vector or normal is CP = .

A vector in the plane we seek is v = . Since the normal is z plane, n $ v = 0.

So,

The equation of the tangent plane is - 3x - 4z - 52 = 0.

Therefore, to find the equation of the tangent plane to a given sphere, dot the radius vector with any vector in the plane, set it equal to zero. We simply write the equation of the plane through point P, with normal vector equal to the vector joining the center of the sphere to the point of tangency.

Practice

1/ a) Find the equation of the sphere with diameter endpoints A (2, 3, 1) and B (-2, 1, 3).

b) Find the equation of the tangent to this sphere at point B.

2/ Find the center and radius of these spheres:

a) x² + y² + z² + 4x - 6z = 0 b) x² + y² + z² - 2x + 4y = 0 c) x² + y² + z² + 6x + 2z + 7 = 0

3/ Write the equations of these spheres: (C is the center point)

a) C (4, 3, 12) through the origin. b) C (4, 3, 12) through (0, 0, 2).

c) C (1, 2, 3) tangent to the xy-plane. d) C (1, - 2, 1) tangent to the z-axis.

e) radius = 2, center in Octant-1, tangent to all 3 coordinate planes.

4/ Find the equation of the tangent plane to x² + y² + z² - 2x - 4y -4 = 0 at P (0, 0, 2).

Solutions

1/ a) C (0, 2, 2), the radius = , so ( x - 0 )² + ( y - 2 )² + ( z - 2 )² = 6

b) Tangent plane has normal n = , and point of tangency is (- 2, 1, 3), so:

The equation of the plane is 2( x + 2 ) + ( y - 1 ) - ( z - 3 ) = 0

2x + y - z + 6 = 0

a) C (-2, 0, 3), radius = b) C (1, - 1, 0), radius = c) C (0, - 3, - 1), radius =

a) ( x - 4 ) ² + ( y - 3) ² + ( z - 12 ) ² = 169 b) ( x - 1 ) ² + ( y - 2) ² + z ² = 9

c) ( x - 1 ) ² + ( y - 2) ² + ( z - 3 ) ² = 9 d) ( x - 1 ) ² + ( y + 2) ² + ( z - 1 ) ² = 5

e) C is (2, 2, 2) u ( x - 2 ) ² + ( y - 2) ² + ( z - 2 ) ² = 4

4/ The center C is (1, 2, 0 ). The radius/normal vector is CP = .

Any vector in the plane through (0, 0, 2) is = . Dot these and set = 0

So,

The equation of this tangent plane is x + 2y - 2z + 4 = 0.

back to Linear Algebra MathRoom Index

a) C (4, 3, 12) through the origin.	b) C (4, 3, 12) through (0, 0, 2).
c) C (1, 2, 3) tangent to the xy-plane.	d) C (1, - 2, 1) tangent to the z-axis.
e) radius = 2, center in Octant-1, tangent to all 3 coordinate planes.

a) ( x - 4 ) ² + ( y - 3) ² + ( z - 12 ) ² = 169	b) ( x - 1 ) ² + ( y - 2) ² + z ² = 9
c) ( x - 1 ) ² + ( y - 2) ² + ( z - 3 ) ² = 9	d) ( x - 1 ) ² + ( y + 2) ² + ( z - 1 ) ² = 5
e) C is (2, 2, 2) u ( x - 2 ) ² + ( y - 2) ² + ( z - 2 ) ² = 4