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Chapter 12: Problem 18

Show that the equation represents a sphere, and ind its center and radius. $$ x^{2}+y^{2}+z^{2}+8 x-6 y+2 z+17=0 $$

Short Answer

Expert verified
The sphere's center is \((-4, 3, -1)\), and the radius is 3.

Step by step solution

01

Identify the Equation's Form

The given equation is \(x^{2} + y^{2} + z^{2} + 8x - 6y + 2z + 17 = 0\). This equation can represent a sphere if it fits into the standard form \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\), where \((h, k, l)\) is the center and \(r\) is the radius.
02

Rearrange to Complete the Square for X

To complete the square for the \(x\)-terms, start with \(x^2 + 8x\). Half of 8 is 4, and 4 squared is 16, so add and subtract 16: \(x^2 + 8x = (x + 4)^2 - 16\).
03

Rearrange to Complete the Square for Y

For the \(y\)-terms, start with \(y^2 - 6y\). Half of -6 is -3, and \(-3\) squared is 9, so add and subtract 9: \(y^2 - 6y = (y - 3)^2 - 9\).
04

Rearrange to Complete the Square for Z

For the \(z\)-terms, start with \(z^2 + 2z\). Half of 2 is 1, and 1 squared is 1, so add and subtract 1: \(z^2 + 2z = (z + 1)^2 - 1\).
05

Rewrite the Equation

Substitute back the squared terms into the original equation: \((x + 4)^2 - 16 + (y - 3)^2 - 9 + (z + 1)^2 - 1 + 17 = 0\).
06

Simplify the Equation

Combine all constants: \(-16 - 9 - 1 + 17 = -9\). Thus, the equation becomes \((x + 4)^2 + (y - 3)^2 + (z + 1)^2 = 9\). This is the standard form of a sphere equation.
07

Identify the Center and Radius

The equation is now in the form \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\). The center is \((-4, 3, -1)\) and the radius is obtained by taking the square root of 9, which is 3.

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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 useful mathematical technique that allows us to rewrite quadratic expressions in a more manageable form. This technique is essential when working with the equation of a sphere, particularly when putting it into its standard form. Each variable term is completed separately. Here's how it works:
  • Divide the Linear Term: Start by taking the linear coefficient (the number next to the variable) and halve it.
  • Square it: Next, square the result. This gives you the number to add and subtract, ensuring the expression fits as a perfect square trinomial.
  • Regroup: Write the squared terms back into the equation and adjust the equation by adding and subtracting the squared number.
For example, in the exercise, we completed the squares for the terms in x, y, and z. For x, it was rearranging from the form \(x^2 + 8x\) to \((x + 4)^2 - 16\). This transformation is necessary to contribute to expressing the equation of a sphere in a simpler form.
Standard Form of a Sphere
The equation of a sphere in standard form is key to understanding its geometric properties. This form is expressed as \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\), where \((h, k, l)\) represents the center coordinates of the sphere, and \(r\) is the radius.
  • Why is it Useful? It's an efficient way to determine both the position and size of a sphere in three-dimensional space.
  • Linking to the Exercise: Our goal was to transform the given equation into this form. By completing the squares and simplifying, we converted it to \((x + 4)^2 + (y - 3)^2 + (z + 1)^2 = 9\), revealing the sphere's details.
The standard form makes it easy to visualize the sphere's structure, akin to identifying a circle in two dimensions. It provides instant insights into its fundamental characteristics.
Center and Radius of a Sphere
Identifying the center and radius of a sphere from its equation is straightforward once the equation is in standard form. From the standard form \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\), you can directly read off these details:
  • Center: The center \(h, k, l\) is derived by changing the signs of numbers inside the parentheses. For \((x + 4)\), it translates to \(x - (-4)\), indicating the x-coordinate of the center is -4.
  • Radius: The radius \(r\) is obtained by taking the square root of the constant term on the right. If the equation is set equal to 9, as in \((x + 4)^2 + (y - 3)^2 + (z + 1)^2 = 9\), then \(r\) is 3.
These values let you pinpoint the sphere's location and determine its size. By identifying the center and radius, you acquire a comprehensive understanding of the sphere's geometry, crucial for further calculations and application.

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Most popular questions from this chapter

\(45-47\) Find the point at which the line intersects the given plane. $$ x=2-2 t, \quad y=3 t, \quad z=1+t ; \quad x+2 y-z=7 $$

(a) Let \(P\) be a point not on the plane that passes through the points \(Q, R,\) and \(S .\) Show that the distance \(d\) from \(P\) to the plane is \(d=\frac{|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|}{|\mathbf{a} \times \mathbf{b}|}\) where \(\mathbf{a}=\overrightarrow{Q R}, \mathbf{b}=\overrightarrow{Q S}\), and \(\mathbf{c}=\overrightarrow{Q P}\) (b) Use the formula in part (a) to find the distance from the point \(P(2,1,4)\) to the plane through the points \(Q(1,0,0)\) \(R(0,2,0),\) and \(S(0,0,3)\)

Which of the following four lines are parallel? Are any of them identical? \(L_{1}: x=1+6 t, \quad y=1-3 t, \quad z=12 t+5\) \(L_{2}: x=1+2 t, \quad y=t, \quad z=1+4 t\) \(L_{3}: 2 x-2=4-4 y=z+1\) \(L_{4}: \mathbf{r}=\langle 3,1,5\rangle+ t\langle 4,2,8\rangle\)

Use intercepts to help sketch the plane. $$ 3 x+y+2 z=6 $$

Show that the curve of intersection of the surfaces \(x^{2}+2 y^{2}-z^{2}+3 x=1\) and \(2 x^{2}+4 y^{2}-2 z^{2}-5 y=0\) lies in a plane.
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