To find the center and radius of the sphere given by the equation x² + y² + z² – 6y – 8z = 0, we’ll first rewrite it in the standard form of a sphere’s equation.
The standard form of a sphere’s equation is (x – h)² + (y – k)² + (z – l)² = r², where (h, k, l) is the center of the sphere and r is the radius.
Let’s start by rearranging the equation:
1. Move the constant to the other side:
x² + y² + z² - 6y - 8z = 0
becomes
x² + y² + z² - 6y - 8z = 0
2. Group terms for y and z:
x² + (y² - 6y) + (z² - 8z) = 0
3. Complete the square for the y and z terms:
For y, take the coefficient of y, which is -6, halve it to get -3, and square it to get 9:
y² - 6y = (y - 3)² - 9
For z, take the coefficient of z, which is -8, halve it to get -4, and square it to get 16:
z² - 8z = (z - 4)² - 16
4. Substitute these back into the equation:
x² + [(y - 3)² - 9] + [(z - 4)² - 16] = 0
This simplifies to:
x² + (y - 3)² + (z - 4)² - 25 = 0
5. Now, move -25 to the right side:
x² + (y - 3)² + (z - 4)² = 25
This equation is now in the standard form of a sphere’s equation.
From this, we can identify:
- Center: (h, k, l) = (0, 3, 4)
- Radius: r = √25 = 5
In conclusion, the center of the sphere is (0, 3, 4) and the radius is 5.