Equation of a Sphere with the Extremities of Diameter Being Given (Cartesian Form)

Understanding the equation of a sphere is a fundamental concept in 3D geometry. In Cartesian coordinates, a sphere can be defined by an equation that relates the coordinates of any point on the sphere to its center and radius. When the extremities of a diameter are given, we can use these points to derive the equation of the sphere.

General Equation of a Sphere

The general equation of a sphere in Cartesian coordinates is given by:

[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 ]

where ((h, k, l)) is the center of the sphere and (r) is the radius.

Deriving the Equation from Diameter Extremities

When the extremities of a diameter are given, we can find the center and the radius of the sphere. Let's denote the extremities of the diameter as (A(x_1, y_1, z_1)) and (B(x_2, y_2, z_2)).

Finding the Center

The center of the sphere, (C(h, k, l)), is the midpoint of the line segment AB. Therefore, the coordinates of the center are:

[ h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2}, \quad l = \frac{z_1 + z_2}{2} ]

Finding the Radius

The radius of the sphere is half the length of the diameter. We can calculate the distance between (A) and (B) using the distance formula and then divide by 2 to get the radius:

[ r = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} ]

Equation of the Sphere

Substituting the center ((h, k, l)) and the radius (r) into the general equation of a sphere, we get the specific equation for the sphere with the given diameter extremities:

[ \left(x - \frac{x_1 + x_2}{2}\right)^2 + \left(y - \frac{y_1 + y_2}{2}\right)^2 + \left(z - \frac{z_1 + z_2}{2}\right)^2 = \left(\frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\right)^2 ]

Important Points and Differences

Point	Description
Center	The center is the midpoint of the diameter's extremities.
Radius	The radius is half the distance between the diameter's extremities.
Equation	The equation incorporates the center and radius to define all points on the sphere.

Example

Let's consider an example to illustrate the process:

Suppose we are given the extremities of a diameter of a sphere as (A(2, -3, 4)) and (B(-4, 5, -6)).

Finding the Center

[ h = \frac{2 + (-4)}{2} = -1, \quad k = \frac{-3 + 5}{2} = 1, \quad l = \frac{4 + (-6)}{2} = -1 ]

So, the center (C) is ((-1, 1, -1)).

Finding the Radius

[ r = \frac{1}{2} \sqrt{(-4 - 2)^2 + (5 - (-3))^2 + (-6 - 4)^2} = \frac{1}{2} \sqrt{36 + 64 + 100} = \frac{1}{2} \sqrt{200} = 5 ]

Equation of the Sphere

[ \left(x + 1\right)^2 + \left(y - 1\right)^2 + \left(z + 1\right)^2 = 5^2 ]

[ x^2 + 2x + 1 + y^2 - 2y + 1 + z^2 + 2z + 1 = 25 ]

[ x^2 + y^2 + z^2 + 2x - 2y + 2z - 22 = 0 ]

This is the equation of the sphere with the given diameter extremities (A) and (B).

By following these steps, you can derive the equation of a sphere in Cartesian form when the extremities of the diameter are given. This knowledge is essential for solving problems in 3D geometry, especially in the context of exams.