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Magazine Chapter 9: Problem 17
Show that the equation represents a sphere, and find its center and radius. $$x^{2}+y^{2}+z^{2}=12 x+2 y$$
Short Answer
Expert verified
The equation represents a sphere with center (6, 1, 0) and radius \(\sqrt{37}\).
Step by step solution
01
Identify the Equation Form
The given equation is \(x^{2}+y^{2}+z^{2}=12x+2y\). This resembles the general form of a sphere equation: \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\). We need to rewrite the given equation in this form.
02
Rearrange the Equation
Move all terms to one side of the equation to set it equal to zero: \(x^2 - 12x + y^2 - 2y + z^2 = 0\).
03
Complete the Square for x-terms
Focus on \(x^2 - 12x\). To complete the square, take half of \(-12\), which is \(-6\), and square it, adding \(36\). This transforms to \((x-6)^2 - 36\).
04
Complete the Square for y-terms
For \(y^2 - 2y\), take half of \(-2\), which is \(-1\), and square it, adding \(1\). This gives \((y-1)^2 - 1\).
05
Rewrite Equation with Completed Squares
Substitute the completed squares back into the equation: \((x-6)^2 - 36 + (y-1)^2 - 1 + z^2 = 0\).
06
Simplify the Equation
Combine the constant terms: \((x-6)^2 + (y-1)^2 + z^2 = 37\). Now the equation is in the standard form of a sphere.
07
Identify Center and Radius
The equation \((x-6)^2 + (y-1)^2 + z^2 = 37\) reveals the center of the sphere at \((6, 1, 0)\) and the radius as \(\sqrt{37}\).
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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 equation into a perfect square trinomial. This technique is very useful for rewriting the equation of a sphere into its standard form. When we apply completing the square in the context of a sphere, the goal is to arrange the terms of each variable into a squared binomial.
To complete the square, follow these steps:
To complete the square, follow these steps:
- Identify and isolate the terms of the variable you are focusing on. For example, if focusing on the x-terms in the equation, they might be grouped as something like \(x^2 - 12x\).
- Take the coefficient of the linear term (the term without a square), divide it by 2, and then square it. This squared value is what you add and subtract within the equation to complete the square. For \(-12x\), you take half of \(-12\) to get \(-6\), squaring it gives \(36\).
- Rewrite the original terms as a binomial squared, like \((x-6)^2\), while remembering to also subtract the squared value you added to balance the equation. In this way, \(x^2 - 12x\) becomes \((x-6)^2 - 36\).
Standard Form of a Sphere
The standard form of a sphere equation is a key concept, as it clearly defines the parameters of the sphere such as its center and radius. This form looks like \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\). Each of these binomial terms squared represents the respective x, y, and z coordinates of the sphere's center.
- The variables \(a\), \(b\), and \(c\) represent the x, y, and z coordinates of the sphere's center in 3-dimensional space.
- The right side of the equation, \(r^2\), represents the radius squared of the sphere.
Center and Radius of a Sphere
Identifying the center and the radius of a sphere from its equation in standard form is straightforward. Once the equation is in the standard form, \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\), you can directly read the center and radius from it.
- The center is located at the point \((a, b, c)\), which you determine from the terms inside the parentheses next to each variable. For example, \((x-6)^2 + (y-1)^2 + z^2\) has a center at \((6, 1, 0)\).
- The radius \(r\) is found by taking the square root of the number on the right side of the equation, which is \(r^2\). Thus, with the equation being equal to 37, the radius \(r\) is \(\sqrt{37}\).
