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Chapter 12: Problem 57

Find the center \(C\) and the radius \(a\) for the spheres in Exercises \(51-60\) $$2 x^{2}+2 y^{2}+2 z^{2}+x+y+z=9$$

Short Answer

Expert verified
The center is \((-\frac{1}{4}, -\frac{1}{4}, -\frac{1}{4})\) and the radius is \(\frac{\sqrt{75}}{4}\).

Step by step solution

01

Identify Sphere Equation

Recognize that the given equation can represent a sphere. A standard sphere equation is given by \((x - h)^2 + (y - k)^2 + (z - l)^2 = a^2\), where \((h, k, l)\) is the center of the sphere and \(a\) is the radius.
02

Simplify the Given Equation

Divide every term of the given equation \(2x^2 + 2y^2 + 2z^2 + x + y + z = 9\) by 2 to simplify it. This gives \(x^2 + y^2 + z^2 + \frac{1}{2}x + \frac{1}{2}y + \frac{1}{2}z = \frac{9}{2}\).
03

Complete the Square

To find the center, complete the square for each variable. Rearrange the terms: \(x^2 + \frac{1}{2}x\), \(y^2 + \frac{1}{2}y\), and \(z^2 + \frac{1}{2}z\). Complete the square:1. For \(x^2 + \frac{1}{2}x\), add and subtract \(\left(\frac{1}{4}\right)^2\): \( (x + \frac{1}{4})^2 - \frac{1}{16}\).2. For \(y^2 + \frac{1}{2}y\), add and subtract \(\left(\frac{1}{4}\right)^2\): \( (y + \frac{1}{4})^2 - \frac{1}{16}\).3. For \(z^2 + \frac{1}{2}z\), add and subtract \(\left(\frac{1}{4}\right)^2\): \( (z + \frac{1}{4})^2 - \frac{1}{16}\).
04

Form the Completed Square Equation

Combine all the completed squares: \((x + \frac{1}{4})^2 + (y + \frac{1}{4})^2 + (z + \frac{1}{4})^2 = \frac{9}{2} + 3 \times \frac{1}{16}\).Simplify the right side by finding a common denominator:\(\frac{9}{2} + \frac{3}{16} = \frac{72}{16} + \frac{3}{16} = \frac{75}{16}\).
05

Identify the Center and Radius

From the equation \((x + \frac{1}{4})^2 + (y + \frac{1}{4})^2 + (z + \frac{1}{4})^2 = \frac{75}{16}\), the center of the sphere is \((-\frac{1}{4}, -\frac{1}{4}, -\frac{1}{4})\). The radius is the square root of the right-hand side: \(a = \sqrt{\frac{75}{16}} = \frac{\sqrt{75}}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This is particularly useful when you need to identify the center and radius of a sphere from its equation.
When dealing with a quadratic expression like \(x^2 + bx\), the goal is to add and subtract a certain value to make it a perfect square trinomial. Let's see how this is done:
  • Identify the coefficient of the linear term, \(b\). In our example, for the term \(x^2 + \frac{1}{2}x\), \(b\) is \(\frac{1}{2}\).
  • Divide this coefficient by 2, which gives \(\frac{1}{4}\). Then you square this result to get \(\frac{1}{16}\).
  • Re-arrange the quadratic as \((x + \frac{1}{4})^2 - \frac{1}{16}\), effectively completing the square for \(x\).
This process is repeated for each variable in the equation. If done correctly, it allows the quadratic expression to be written in a form that makes it easier to derive the sphere's center and radius.
Center of a Sphere
The center of a sphere is a crucial point that helps in understanding its geometric properties. When the equation of a sphere is put in its standard form, \((x - h)^2 + (y - k)^2 + (z - l)^2 = a^2\), the center \(C\) can be identified as the point \((h, k, l)\).
In the given exercise, after simplifying and completing the square, we obtain the equation:
  • \((x + \frac{1}{4})^2 + (y + \frac{1}{4})^2 + (z + \frac{1}{4})^2 = \frac{75}{16}\).
To find the center, look for the values that make each bracket equal zero:
  • The subtraction of a positive \(\frac{1}{4}\) from each variable indicates that the center coordinates are negatives of these terms.
  • Thus, the center of the sphere is \((-\frac{1}{4}, -\frac{1}{4}, -\frac{1}{4})\).
Radius of a Sphere
The radius of a sphere determines its size and is derived from the standard sphere equation, \((x - h)^2 + (y - k)^2 + (z - l)^2 = a^2\). It is the distance from the center of the sphere to any point on its surface.
From our derived equation, \((x + \frac{1}{4})^2 + (y + \frac{1}{4})^2 + (z + \frac{1}{4})^2 = \frac{75}{16}\), the term on the right-hand side corresponds to \(a^2\).
  • To find the radius, take the square root of \(\frac{75}{16}\).
  • The calculation is: \(a = \sqrt{\frac{75}{16}}\).
  • Thus, the radius of the sphere is \(\frac{\sqrt{75}}{4}\).
Understanding the radius is paramount as it directly affects the sphere's proportion and spatial properties.

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Most popular questions from this chapter

In Exercises \(17-24\) , describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities. \begin{equation} { a. }1 \leq x^{2}+y^{2}+z^{2} \leq 4 \quad \text { b. } x^{2}+y^{2}+z^{2} \leq 1, \quad z \geq 0 \end{equation}

Double cancelation If \(\mathbf { u } \neq \mathbf { 0 }\) and if \(\mathbf { u } \times \mathbf { v } = \mathbf { u } \times \mathbf { w }\) and \(\mathbf { u } \cdot \mathbf { v } = \mathbf { u } \cdot \mathbf { w } ,\) then does \(\mathbf { v } = \mathbf { w } ?\) Give reasons for your answer.

Find the distance from the point \((3,-4,2)\) to the \begin{equation} \text { a. }x y \text { -plane } \quad \text { b. } y z \text { -plane } \quad \text { c. } x z \end{equation}

In Exercises \(17-24\) , describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities. \begin{equation} \text {a. }x \geq 0, \quad y \geq 0, \quad z=0 \quad \text { b. } x \geq 0, \quad y \leq 0, \quad z=0 \end{equation}

Use a CAS to plot the surfaces in Exercises \(53-58 .\) Identify the type of quadric surface from your graph. $$ y-\sqrt{4-z^{2}}=0 $$
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