Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 12: Problem 15
\(15-18\) Show that the equation represents a sphere, and find its center and radius. $$x^{2}+y^{2}+z^{2}-6 x+4 y-2 z=11$$
Short Answer
Expert verified
The sphere's center is (3, -2, 1) and its radius is 5.
Step by step solution
01
Identify the general form of a sphere equation
A sphere's equation in general form is \( (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \), where \((h, k, l)\) is the center and \(r\) is the radius.
02
Complete the square for the x-terms
Rewrite the \(x\)-related terms: \(x^2 - 6x\). To complete the square, add and subtract \(\left(\frac{-6}{2}\right)^2 = 9\). This gives \(\left(x^2 - 6x + 9\right) - 9\), which simplifies to \((x - 3)^2 - 9\).
03
Complete the square for the y-terms
Rewrite the \(y\)-related terms: \(y^2 + 4y\). To complete the square, add and subtract \(\left(\frac{4}{2}\right)^2 = 4\). This gives \(\left(y^2 + 4y + 4\right) - 4\), which simplifies to \((y + 2)^2 - 4\).
04
Complete the square for the z-terms
Rewrite the \(z\)-related terms: \(z^2 - 2z\). To complete the square, add and subtract \(\left(\frac{-2}{2}\right)^2 = 1\). This gives \(\left(z^2 - 2z + 1\right) - 1\), which simplifies to \((z - 1)^2 - 1\).
05
Substitute completed squares into the original equation
Substitute the completed square terms into the original equation: \((x - 3)^2 - 9 + (y + 2)^2 - 4 + (z - 1)^2 - 1 = 11\).
06
Simplify the equation
Combine all constant terms: \((x - 3)^2 + (y + 2)^2 + (z - 1)^2 = 11 + 9 + 4 + 1\). Simplifying gives \((x - 3)^2 + (y + 2)^2 + (z - 1)^2 = 25\).
07
State the center and the radius
The equation \((x - 3)^2 + (y + 2)^2 + (z - 1)^2 = 25\) represents a sphere. The center is \((3, -2, 1)\) and the radius is \( \sqrt{25} = 5 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
To find the equation of a sphere in a recognizable form, we often use a method called **Completing the Square**. This technique involves rewriting quadratic expressions so they resemble perfect squares. Let's break it down step-by-step for clarity:
- Consider the equation: \(x^2 - 6x\). Try to rewrite it as a perfect square.
- First, take half of the coefficient of \(x\) (i.e., -6), divide it by 2 to get -3, then square it to get 9.
- Add and subtract 9 within the equation. Group the terms within parentheses: \((x - 3)^2 - 9\).
- For \(y^2 + 4y\), add and subtract 4 to get \((y + 2)^2 - 4\).
- For \(z^2 - 2z\), add and subtract 1 to get \((z - 1)^2 - 1\).
Center of a Sphere
Once we have rewritten the quadratic parts of the equation using the technique of completing the square, we achieve a form that resembles this: \[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\]In this form, the center of the sphere is denoted by the coordinates \((h, k, l)\). From our example, by bringing all terms together:
The orderly capture of these constants is essential for further calculations involving the sphere.
- \((x - 3)^2\) suggests the center's x-coordinate is 3.
- \((y + 2)^2\) translates to a y-coordinate of -2 (since the form is \((y-k)^2\)),
- \((z - 1)^2\) gives the z-coordinate as 1.
The orderly capture of these constants is essential for further calculations involving the sphere.
Radius of a Sphere
Having identified the sphere's center, the next step is to find its radius. The equation obtained after completing the square is: \[(x-3)^2 + (y+2)^2 + (z-1)^2 = 25\]The term on the right side of the equation, \(25\), represents \(r^2\), where \(r\) is the radius of the sphere. To find the radius, take the square root of this term.
- Therefore, \(r^2 = 25\) leads to \(r = \sqrt{25}\).
- Simplifying \(\sqrt{25}\) gives us a radius \(r = 5\).
