The problem asks us to find the center and radius of the sphere given by the equation $4x^2 + 4y^2 + 4z^2 - 4x + 8y + 16z - 13 = 0$ by completing the square.

1. Problem Description

The problem asks us to find the center and radius of the sphere given by the equation $4x^2 + 4y^2 + 4z^2 - 4x + 8y + 16z - 13 = 0$ by completing the square.

2. Solution Steps

First, divide the entire equation by 4:

$x^2 + y^2 + z^2 - x + 2y + 4z - \frac{13}{4} = 0$

Now, we group the $x$, $y$, and $z$ terms together:

$(x^2 - x) + (y^2 + 2y) + (z^2 + 4z) = \frac{13}{4}$

Complete the square for each group:

For $x^2 - x$, we need to add and subtract $(\frac{-1}{2})^2 = \frac{1}{4}$.

For $y^2 + 2y$, we need to add and subtract $(\frac{2}{2})^2 = 1$.

For $z^2 + 4z$, we need to add and subtract $(\frac{4}{2})^2 = 4$.

So, we have:

$(x^2 - x + \frac{1}{4} - \frac{1}{4}) + (y^2 + 2y + 1 - 1) + (z^2 + 4z + 4 - 4) = \frac{13}{4}$

$(x - \frac{1}{2})^2 - \frac{1}{4} + (y + 1)^2 - 1 + (z + 2)^2 - 4 = \frac{13}{4}$

Move the constants to the right side:

$(x - \frac{1}{2})^2 + (y + 1)^2 + (z + 2)^2 = \frac{13}{4} + \frac{1}{4} + 1 + 4$

$(x - \frac{1}{2})^2 + (y + 1)^2 + (z + 2)^2 = \frac{14}{4} + 5 = \frac{7}{2} + \frac{10}{2} = \frac{17}{2}$

The equation of a sphere is given by $(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$, where $(a, b, c)$ is the center and $r$ is the radius.

Comparing this to our equation, we have:

Center: $(\frac{1}{2}, -1, -2)$

$r^2 = \frac{17}{2}$, so $r = \sqrt{\frac{17}{2}} = \frac{\sqrt{34}}{2}$

3. Final Answer

Center: $(\frac{1}{2}, -1, -2)$

Radius: $\frac{\sqrt{34}}{2}$