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Magazine Chapter 12: Problem 59
Find the center \(C\) and the radius \(a\) for the spheres in Exercises \(51-60\) $$x^{2}+y^{2}+z^{2}-4 x+6 y-10 z=11$$
Short Answer
Expert verified
The center is at \((2, -3, 5)\) and the radius is 7.
Step by step solution
01
Identify the sphere equation
The given equation is a general form for a sphere: \[ x^2 + y^2 + z^2 - 4x + 6y - 10z = 11 \]We need to rewrite this equation to identify the center and radius of the sphere by completing the square for each variable.
02
Complete the square for the x terms
For the terms involving \(x\), which are \(x^2 - 4x\), complete the square:\[ x^2 - 4x = (x - 2)^2 - 4 \]This represents a square with center at \(x = 2\) and an added constant \(-4\).
03
Complete the square for the y terms
For the terms involving \(y\), which are \(y^2 + 6y\), complete the square:\[ y^2 + 6y = (y + 3)^2 - 9 \]This represents a square with center at \(y = -3\) and an added constant \(-9\).
04
Complete the square for the z terms
For the terms involving \(z\), which are \(z^2 - 10z\), complete the square:\[ z^2 - 10z = (z - 5)^2 - 25 \]This represents a square with center at \(z = 5\) and an added constant \(-25\).
05
Rewrite the equation of the sphere
Substitute the completed squares back into the original equation:\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 + (z - 5)^2 - 25 = 11 \]Combine the constants to simplify the equation:\[ (x - 2)^2 + (y + 3)^2 + (z - 5)^2 = 49 \]This is the standard form of the sphere's equation.
06
Determine the center and radius
From the equation \((x-2)^2 + (y+3)^2 + (z-5)^2 = 49\), identify the center and radius:- The center \(C\) is at the point \((2, -3, 5)\).- The radius \(a\) is \(\sqrt{49} = 7\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of a Sphere
The equation of a sphere in its standard form is essential to understanding its position and size in space. The general equation for a sphere is \[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]where
- \((h, k, l)\) is the center of the sphere.
- \(r\) is the radius of the sphere.
Completing the Square
To convert a general quadratic equation into a system that displays the center and radius of a sphere, we use a method called "completing the square." This procedure involves turning a quadratic expression of each variable into a perfect square trinomial.
Suppose you have a quadratic term like \( x^2 - 4x \).
- You take the coefficient of \(x\),divide it by 2, and square the result.
This gives \((\frac{-4}{2})^2 = (-2)^2 = 4\).
Then rewrite the expression as a perfect square minus a constant: \( (x-2)^2 - 4 \).
By repeating this process for the \(y\) and \(z\) terms, we can transform the complex equation into a format that reveals the sphere's structure.
This makes completing the square an invaluable tool when dealing with spheres.
Suppose you have a quadratic term like \( x^2 - 4x \).
- You take the coefficient of \(x\),divide it by 2, and square the result.
This gives \((\frac{-4}{2})^2 = (-2)^2 = 4\).
Then rewrite the expression as a perfect square minus a constant: \( (x-2)^2 - 4 \).
By repeating this process for the \(y\) and \(z\) terms, we can transform the complex equation into a format that reveals the sphere's structure.
This makes completing the square an invaluable tool when dealing with spheres.
Center of a Sphere
The center of a sphere is one of the most critical features. It is the point from which every point on the surface of the sphere is equidistant.
In the completed square equation \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \],
we identify the center as the coordinates \((h, k, l)\).
Following the conversion of our given equation \( x^2 + y^2 + z^2 - 4x + 6y - 10z = 11 \)into its new form,
the center is found to be \( (2, -3, 5) \).
This information is pivotal since knowing the center allows us to anchor the sphere's position within three-dimensional space.
In the completed square equation \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \],
we identify the center as the coordinates \((h, k, l)\).
Following the conversion of our given equation \( x^2 + y^2 + z^2 - 4x + 6y - 10z = 11 \)into its new form,
the center is found to be \( (2, -3, 5) \).
This information is pivotal since knowing the center allows us to anchor the sphere's position within three-dimensional space.
Radius of a Sphere
The radius of a sphere is the constant distance from the center to any point on its outer surface. In the standard form of a sphere's equation,\( (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \),
\(r^2\) denotes the radius squared. For the sphere we examined, the equation simplifies to\[(x - 2)^2 + (y + 3)^2 + (z - 5)^2 = 49\].
Thus, the radius \(r\) is determined by taking the square root of \(49\),
which results in a radius of \(7\).
This gives us a sense of the sphere's size, allowing one to visualize or calculate other attributes related to its space and geometry.
\(r^2\) denotes the radius squared. For the sphere we examined, the equation simplifies to\[(x - 2)^2 + (y + 3)^2 + (z - 5)^2 = 49\].
Thus, the radius \(r\) is determined by taking the square root of \(49\),
which results in a radius of \(7\).
This gives us a sense of the sphere's size, allowing one to visualize or calculate other attributes related to its space and geometry.
