Question

Complete the squares to find the center and radius of the sphere whose equation is given (see Example 2). \(x^{2}+y^{2}+z^{2}+2 x-6 y-10 z+34=0\)

Short answer

The center is at (-1, 3, 5) and the radius is 1.

Step-by-step solution

Group Terms

We start by arranging the given equation \[ x^{2}+y^{2}+z^{2}+2x-6y-10z+34=0 \]such that all the variables that are related in the same square are grouped together: \[ (x^{2} + 2x) + (y^{2} - 6y) + (z^{2} - 10z) + 34 = 0 \]

Complete the Square for x

Take the terms relating to \( x \): \[ x^{2} + 2x \]To complete the square, take half of the coefficient of \( x \), square it, and add and subtract it inside the bracket. \[ x^{2} + 2x = (x^{2} + 2x + 1) - 1 \]This simplifies to: \[ (x + 1)^{2} - 1 \]

Complete the Square for y

Next, consider the \( y \) terms: \[ y^{2} - 6y \]Complete the square by taking half of -6 (which is -3), squaring it, and adding and subtracting:\[ y^{2} - 6y = (y^{2} - 6y + 9) - 9 \]This becomes: \[ (y - 3)^{2} - 9 \]

Complete the Square for z

For the \( z \) variable: \[ z^{2} - 10z \]Take half of -10, square it, then add and subtract that value:\[ z^{2} - 10z = (z^{2} - 10z + 25) - 25 \]This simplifies to:\[ (z - 5)^{2} - 25 \]

Substitute Completed Squares Back

Substituting the completed squared forms back into the equation, we have:\[ ((x + 1)^{2} - 1) + ((y - 3)^{2} - 9) + ((z - 5)^{2} - 25) + 34 = 0 \]

Simplify the Equation

Simplify by combining all the constants:\[ (x + 1)^{2} + (y - 3)^{2} + (z - 5)^{2} - 1 - 9 - 25 + 34 = 0 \]\[ (x + 1)^{2} + (y - 3)^{2} + (z - 5)^{2} - 1 = 0 \]Finally adding 1 to both sides, we get: \[ (x + 1)^{2} + (y - 3)^{2} + (z - 5)^{2} = 1 \]

Identify the Center and the Radius

The standard form of the sphere's equation is \[ (x - h)^{2} + (y - k)^{2} + (z - l)^{2} = r^{2} \],where \( (h, k, l) \) is the center and \( r \) is the radius. From our equation \[ (x + 1)^{2} + (y - 3)^{2} + (z - 5)^{2} = 1 \],we can see that the center is \( (-1, 3, 5) \) and the radius \( r \) is 1.

Key concepts

Completing the Square

Completing the square is a fundamental algebraic technique used to manipulate quadratic expressions into a perfect square trinomial. This method transforms a quadratic expression, making it easier to analyze or graph. Let's see how it works with an example from the equation of a sphere.
For instance, consider the terms relating to the variable \( x \) in the given equation:

• Start with: \( x^{2} + 2x \).
• To transform this into a perfect square trinomial, take half of the coefficient of \( x \) (which is 2), then square it. This results in 1.
• Add and subtract this squared value inside the expression: \( x^{2} + 2x = (x + 1)^{2} - 1 \).

By repeating this process with each variable component in the equation, you can remodel the quadratic part of the equation into a series of perfect squares. This makes it much simpler to identify features like the center and radius of a sphere, which is essential in various mathematical and practical applications such as geometry and physics.

Center of a Sphere

The center of a sphere is one of its defining characteristics in its geometric equation. In mathematical terms, this is known as the coordinate \( (h, k, l) \) in the equation of a sphere:
\[(x - h)^{2} + (y - k)^{2} + (z - l)^{2} = r^{2}\]
Once the equation is in this standard form, identifying the center is straightforward. In our completed equation:
\[(x + 1)^{2} + (y - 3)^{2} + (z - 5)^{2} = 1\]
you can observe:

• The expression \((x + 1)^{2}\) relates to \( (x - (-1))^{2} \), which indicates the x-coordinate of the center is -1.
• The expression \((y - 3)^{2}\) gives the y-coordinate, 3.
• Finally, \((z - 5)^{2}\) reveals the z-coordinate as 5.

Hence, the center of the sphere is at the point \((-1, 3, 5)\). Finding the center is crucial for understanding the sphere's position in a three-dimensional space, which is especially useful in fields like computer graphics, engineering, and beyond.

Radius of a Sphere

The radius of a sphere is the distance from its center to any point on its surface. It represents one of the key dimensions of a sphere.
In the standard equation of a sphere:
\[(x - h)^{2} + (y - k)^{2} + (z - l)^{2} = r^{2}\]

• The term \( r^{2} \) on the right-hand side is the square of the radius.
• From the equation of our sphere \((x + 1)^{2} + (y - 3)^{2} + (z - 5)^{2} = 1\), we can clearly see \( r^{2} = 1 \).
• Solving for \( r \), take the square root of both sides, yielding \( r = \sqrt{1} = 1 \).

Understanding the radius is critical for solving problems related to volume and surface area, among others. In practical applications like navigation and satellite technology, knowing a sphere's radius can be pivotal in calculating distances and designing pathways.