Obtain the equation of the sphere with the points `(1, -1, 1) and (3, -3, 3)` as the extremities of a diametre and find the coordinate of its centre.

Answer

To obtain the equation of the sphere with the given points \((1, -1, 1)\) and \((3, -3, 3)\) as the extremities of a diameter, we can follow these steps:

### Step 1: Find the center of the sphere
The center of the sphere is the midpoint of the diameter. The midpoint \(M\) of two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is given by the formula:

\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]

For our points \((1, -1, 1)\) and \((3, -3, 3)\):

\[ M = \left( \frac{1 + 3}{2}, \frac{-1 - 3}{2}, \frac{1 + 3}{2} \right) = \left( \frac{4}{2}, \frac{-4}{2}, \frac{4}{2} \right) = (2, -2, 2) \]

### Step 2: Calculate the radius of the sphere
The radius \(r\) of the sphere is the distance from the center to either of the endpoints of the diameter. We can use the distance formula between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\):

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

Using the center \((2, -2, 2)\) and one endpoint \((1, -1, 1)\):

\[ r = \sqrt{(1 - 2)^2 + (-1 + 2)^2 + (1 - 2)^2} = \sqrt{(-1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 3: Write the equation of the sphere
The equation of a sphere with center \((h, k, l)\) and radius \(r\) is given by:

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

Substituting \(h = 2\), \(k = -2\), \(l = 2\), and \(r = \sqrt{3}\):

\[ (x - 2)^2 + (y + 2)^2 + (z - 2)^2 = (\sqrt{3})^2 \]

This simplifies to:

\[ (x - 2)^2 + (y + 2)^2 + (z - 2)^2 = 3 \]

### Final Answer
The equation of the sphere is:

\[ (x - 2)^2 + (y + 2)^2 + (z - 2)^2 = 3 \]

The coordinates of the center of the sphere are \((2, -2, 2)\).

---