Question

Find the equation of the sphere centered at (-9,9, -9) with radius 5. Normalize your equations so that the coefficient of x2 is 1. -0.
Give an equation which describes the intersection of this sphere with the plane z = 0.

Answer 1

Answer:

(a) x² + y² + z² + 18(x - y + z) + 218 = 0

(b) (x + 9)² + (y - 9)² + 56 = 0

Step-by-step explanation:

The general equation of a sphere of radius r and centered at C = (x₀, y₀, z₀) is given by;

(x - x₀)² + (y - y₀)² + (z - z₀)² = r²               ------------------(i)

From the question:

The sphere is centered at C = (x₀, y₀, z₀) = (-9, 9, -9) and has a radius r = 5.

Therefore, to get the equation of the sphere, substitute these values into equation (i) as follows;

(x - (-9))² + (y - 9)² + (z - (-9))² = 5²

(x + 9)² + (y - 9)² + (z + 9)² = 25      ------------------(ii)

Open the brackets and have the following:

(x + 9)² + (y - 9)² + (z + 9)² = 25

(x² + 18x + 81) + (y² - 18y + 81) + (z² + 18z + 81) = 25

x² + 18x + 81 + y² - 18y + 81 + z² + 18z + 81 = 25

x² + y² + z² + 18(x - y + z) + 243 = 25

x² + y² + z² + 18(x - y + z) + 218 = 0    [equation has already been normalized since the coefficient of x² is 1]

Therefore, the equation of the sphere centered at (-9,9, -9) with radius 5 is:

x² + y² + z² + 18(x - y + z) + 218 = 0

(2)  To get the equation when the sphere intersects a plane z = 0, we substitute z = 0 in equation (ii) as follows;

(x + 9)² + (y - 9)² + (0 + 9)² = 25

(x + 9)² + (y - 9)² + (9)² = 25

(x + 9)² + (y - 9)² + 81 = 25        [subtract 25 from both sides]

(x + 9)² + (y - 9)² + 81 - 25 = 25 - 25

(x + 9)² + (y - 9)² + 56 = 0

The equation is therefore, (x + 9)² + (y - 9)² + 56 = 0