Question

Complete the square to write the equation of the sphere in standard form. Find the center and radius. \(9 x^{2}+9 y^{2}+9 z^{2}-6 x+18 y+1=0\)

Short answer

The standard form of the given equation is \((x -\frac{1}{3})^2 + (y + 1)^2 + z^2 = 1\), and the center is at (\(\frac{1}{3}, -1, 0\)) with radius 1.

Step-by-step solution

Group the terms

Group the squared terms, the linear terms and constants separately. The equation can be rewritten as \(9(x^2 -\frac{2}{3}x) + 9(y^2 + 2y) + 9z^2 = -1\). Divide the whole equation by 9 to simplify it into \(x^2 -\frac{2}{3}x + y^2 + 2y + z^2 = - \frac{1}{9}\).

Complete the square for each group

To complete the square, the goal is to make each squared group into a perfect square. Take half of the coefficient of \(x\), and square it, (-2/3)/2 = -1/3, (-1/3) ^ 2 = 1/9. Add this inside the parenthesis to get \((x^2 -\frac{2}{3}x + \frac{1}{9})\). Repeat this for the \(y\) term which is 2/2 = 1, 1^2 = 1; Hence, \((y^2 +2y +1)\). Thus, the equation becomes \((x -\frac{1}{3})^2 + (y + 1)^2 + z^2 = -\frac{1}{9} + \frac{1}{9} + 1\).

Simplify the equation

Lastly, simplify \( -\frac{1}{9} + \frac{1}{9} + 1\) on the right side of the equation to get 1, and leave each group in the form \((variable - value)^2\). The equation in standard form becomes \((x -\frac{1}{3})^2 + (y + 1)^2 + z^2 = 1\).

Find the center and radius of the sphere

The center of the sphere is determined by the values inside the parentheses, and is \((h, k, l) = (\frac{1}{3}, -1, 0)\), and the radius is the square root of the right hand side, which means \(r = \sqrt{1} = 1\).