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Magazine Chapter 12: Problem 28
Find the center and radius of the sphere with the given equation. \(2 x^{2}+2 y^{2}+2 z^{2}=7 x+9 y+11 z\)
Short Answer
Expert verified
The center is \((\frac{7}{4}, \frac{9}{4}, \frac{11}{4})\) and the radius is \(\frac{\sqrt{251}}{4}\).
Step by step solution
01
Rearrange the Equation
Start by rearranging the given equation by moving the terms on the right side to the left side. This gives us: \[2x^2 + 2y^2 + 2z^2 - 7x - 9y - 11z = 0\]
02
Simplify the Equation
Factor out a 2 from the left side of the equation: \[2(x^2 + y^2 + z^2 - \frac{7}{2}x - \frac{9}{2}y - \frac{11}{2}z) = 0\] Divide the whole equation by 2: \[x^2 + y^2 + z^2 - \frac{7}{2}x - \frac{9}{2}y - \frac{11}{2}z = 0\]
03
Complete the Square for x
To complete the square for the x terms, focus on \(x^2 - \frac{7}{2}x\). Add and subtract \(\left(\frac{7}{4}\right)^2\):\[x^2 - \frac{7}{2}x = (x - \frac{7}{4})^2 - (\frac{7}{4})^2\]
04
Complete the Square for y
Now complete the square for the y terms, \(y^2 - \frac{9}{2}y\). Add and subtract \(\left(\frac{9}{4}\right)^2\):\[y^2 - \frac{9}{2}y = (y - \frac{9}{4})^2 - (\frac{9}{4})^2\]
05
Complete the Square for z
Complete the square for the z terms, \(z^2 - \frac{11}{2}z\). Add and subtract \(\left(\frac{11}{4}\right)^2\):\[z^2 - \frac{11}{2}z = (z - \frac{11}{4})^2 - (\frac{11}{4})^2\]
06
Combine and Simplify
Combine all the completed squares into the equation:\[(x - \frac{7}{4})^2 - \frac{49}{16} + (y - \frac{9}{4})^2 - \frac{81}{16} + (z - \frac{11}{4})^2 - \frac{121}{16} = 0\]Simplify:\[(x - \frac{7}{4})^2 + (y - \frac{9}{4})^2 + (z - \frac{11}{4})^2 = \frac{251}{16}\]
07
Identify the Center and Radius
The standard form for a sphere is: \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\).Here, - Center \((h, k, l)\) is \((\frac{7}{4}, \frac{9}{4}, \frac{11}{4})\).- Radius \(r\) is \(\sqrt{\frac{251}{16}}\), which simplifies to \(\frac{\sqrt{251}}{4}\).
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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 mathematical technique used to transform a quadratic equation into a perfect square trinomial. This method is especially useful in shifting equations into a format that can help identify specific geometric properties. For instance, in the context of a sphere equation, it helps in finding the center and the radius.
Here's a step-by-step approach to completing the square, demonstrated with the example of the sphere equation:
Here's a step-by-step approach to completing the square, demonstrated with the example of the sphere equation:
- Focus on each variable one at a time. For the equation given, this means addressing the terms for \(x\), \(y\), and \(z\) individually.
- To complete the square for the \(x\) term \(x^2 - \frac{7}{2}x\), add and subtract \((\frac{7}{4})^2\). This transforms into \((x - \frac{7}{4})^2 - (\frac{7}{4})^2\).
- Repeat this process for the \(y\) term, \(y^2 - \frac{9}{2}y\), by adding and subtracting \((\frac{9}{4})^2\) which becomes \((y - \frac{9}{4})^2 - (\frac{9}{4})^2\).
- Finally, tackle the \(z\) term, \(z^2 - \frac{11}{2}z\), by adding and subtracting \((\frac{11}{4})^2\), resulting in \((z - \frac{11}{4})^2 - (\frac{11}{4})^2\).
- Combine the completed squares and move constants to one side: \((x - \frac{7}{4})^2 + (y - \frac{9}{4})^2 + (z - \frac{11}{4})^2\).
Center of a Sphere
In geometry, the center of a sphere is the point that is equidistant from all points on the surface of the sphere. Determining the center is crucial for understanding the sphere's position in a coordinate space.
From the equation of the sphere in standard form: \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), the point \((h, k, l)\) represents the center.
To identify the center from the given equation:
From the equation of the sphere in standard form: \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), the point \((h, k, l)\) represents the center.
To identify the center from the given equation:
- The coefficients derived from the completed square method, i.e., \((x - \frac{7}{4})\), \((y - \frac{9}{4})\), and \((z - \frac{11}{4})\), lead directly to the center coordinates.
- Here, the center is found at \((\frac{7}{4}, \frac{9}{4}, \frac{11}{4})\). This gives a precise location in space where the center of the sphere lies.
Radius of a Sphere
The radius of a sphere is the distance from its center to any point on its surface. Understanding how to extract this from the standard form of a sphere's equation is fundamental to solving geometrical problems.
In the equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), \(r^2\) represents the square of the radius. This means the actual radius \(r\) is the square root of this term.
In our simplified equation:
In the equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), \(r^2\) represents the square of the radius. This means the actual radius \(r\) is the square root of this term.
In our simplified equation:
- The constant on the right side after completing the square is \(\frac{251}{16}\), which is equal to \(r^2\).
- To find \(r\), compute the square root: \(r = \sqrt{\frac{251}{16}}\).
- This simplifies to \(r = \frac{\sqrt{251}}{4}\).
