Problem 16 - Algebra and Trigonometry

Question

\(15-18=\) Show that the equation represents a sphere, and find its center and radius. $$ x^{2}+y^{2}+z^{2}+4 x-6 y+2 z=10 $$

Step-by-Step Solution

Answer

The sphere's center is at (-2, 3, -1) and its radius is 2√6.

Step 1: Rearrange Equation into Standard Form

First, we need to express the given equation in the standard format of a sphere's equation. A sphere's equation typically has the form \[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\]We start by grouping the quadratic terms with the corresponding linear terms: \[x^2 + 4x + y^2 - 6y + z^2 + 2z = 10\]

Step 2: Complete the Square for X-Terms

To complete the square for the terms involving \(x\), we take \(x^2 + 4x\) and add and subtract \((\frac{4}{2})^2 = 4\):\[(x^2 + 4x) = (x + 2)^2 - 4\]

Step 3: Complete the Square for Y-Terms

For the \(y\) terms, we complete the square using the expression \(y^2 - 6y\): Add and subtract \((\frac{-6}{2})^2 = 9\):\[(y^2 - 6y) = (y - 3)^2 - 9\]

Step 4: Complete the Square for Z-Terms

For the \(z\) terms, complete the square for \(z^2 + 2z\) by adding and subtracting \((\frac{2}{2})^2 = 1\):\[(z^2 + 2z) = (z + 1)^2 - 1\]

Step 5: Combine Perfect Squares

Insert the completed square terms back into the equation:\[(x + 2)^2 - 4 + (y - 3)^2 - 9 + (z + 1)^2 - 1 = 10\]Combine the constants on the left side:\[(x + 2)^2 + (y - 3)^2 + (z + 1)^2 = 10 + 4 + 9 + 1\]

Step 6: Solve for the Right Side and Identify Sphere Parameters

Simplifying the right side:\[(x + 2)^2 + (y - 3)^2 + (z + 1)^2 = 24\]This indicates the sphere has a center at \((-2, 3, -1)\) and a radius of \(\sqrt{24} = 2\sqrt{6}\).

Key Concepts

Completing the Square

Completing the square is a technique used to convert a quadratic expression into a perfect square trinomial. This is particularly useful in rewriting equations into standard forms, such as the equation of a sphere.
To complete the square, you follow several steps:

• Take the quadratic term plus the linear term, for instance, terms like \(x^2 + 4x\).
• Find the square of half of the coefficient of the linear term. Here, \(\left(\frac{4}{2}\right)^2 = 4\).
• Add and subtract this square inside the equation so that the trinomial becomes a perfect square: \((x + 2)^2 - 4\).

This process is repeated for each variable group \((y\) and \(z\)), thereby simplifying each into a perfect square, which is essential for formatting equations into their standard forms, such as that of a sphere.

Standard Form of a Sphere

The equation of a sphere is typically written in its standard form to easily identify its center and radius. The standard form of a sphere's equation is:

• \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\)

Here, \((h, k, l)\) is the center and \(r\) is the radius of the sphere. This form is obtained by applying completing the square method to each variable term in the original equation.
In our case, the steps turn the original polynomial \(x^2 + y^2 + z^2 + 4x - 6y + 2z = 10\) into the standard form \((x + 2)^2 + (y - 3)^2 + (z + 1)^2 = 24\), identifying the geometrical nature of the object represented by the equation.

Center of a Sphere

The center of a sphere in the standard equation is given by the coordinates \((h, k, l)\). These are extracted directly from the transformed equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). For instance, in \((x+2)^2 + (y-3)^2 + (z+1)^2\), we identify the center as \((-2, 3, -1)\).
These values are deduced as being the symmetric opposite of what's within the parenthesis alongside each variable term:

• For \((x + 2)\), the shift is \(-2\).
• For \((y - 3)\), the shift is \(3\).
• For \((z + 1)\), the shift is \(-1\).

The center point \((-2, 3, -1)\) geometrically indicates the exact middle of the sphere, from which all points on its surface are equidistant.

Radius of a Sphere

The radius of the sphere is the distance from its center to any point on its surface. In the sphere's standard form equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), \(r\) represents the radius.
By looking at the equation \((x+2)^2 + (y-3)^2 + (z+1)^2 = 24\), we can deduce that \(r^2 = 24\). By taking the square root, the radius of the sphere becomes \(r = \sqrt{24}\), which simplifies to \(r = 2\sqrt{6}\).

• This transformation reveals the sphere's size in three-dimensional space.
• Simplifying the expression ensures clarity in understanding the sphere's radius.

In essence, the radius \(2\sqrt{6}\) not only represents the size but also serves as a crucial element in defining the spatial properties of the sphere.