Problem 10 - APEX Calculus

Question

Find the center and radius of the sphere defined by \(x^{2}+y^{2}+z^{2}+4 x-2 y-4 z+4=0\)

Step-by-Step Solution

Answer

Center: (-2, 1, 2); Radius: \(\sqrt{5}\).

Step 1: Group and Rearrange Terms

Rearrange the given equation to better identify terms that can be completed into perfect squares. The equation is currently: \[ x^2 + y^2 + z^2 + 4x - 2y - 4z + 4 = 0 \]Reorganize it as follows:\[ (x^2 + 4x) + (y^2 - 2y) + (z^2 - 4z) = -4 \]

Step 2: Complete the Square for x

Perform completing the square on the \(x\)-terms. Take the coefficient of \(x\), divide by 2, square it, and add/subtract it:- Coefficient of \(x\) is 4.- \((4/2)^2 = 4\).So, \(x^2 + 4x\) becomes \( (x^2 + 4x + 4 - 4) \) or \((x+2)^2-4\).

Step 3: Complete the Square for y

Perform completing the square on the \(y\)-terms. Take the coefficient of \(y\), divide by 2, square it, and add/subtract it:- Coefficient of \(y\) is -2.- \((-2/2)^2 = 1\).So, \(y^2 - 2y\) becomes \((y^2 - 2y + 1 - 1)\) or \((y-1)^2-1\).

Step 4: Complete the Square for z

Perform completing the square on the \(z\)-terms. Take the coefficient of \(z\), divide by 2, square it, and add/subtract it:- Coefficient of \(z\) is -4.- \((-4/2)^2 = 4\).So, \(z^2 - 4z\) becomes \((z^2 - 4z + 4 - 4)\) or \((z-2)^2-4\).

Step 5: Write the Complete Squares

Substitute the completed square forms back into the equation:\[((x+2)^2 - 4) + ((y-1)^2 - 1) + ((z-2)^2 - 4) = -4 \]

Step 6: Simplify the Equation

Combine all constants and simplify:\[(x+2)^2 + (y-1)^2 + (z-2)^2 - 9 = -4 \]Add 9 to both sides:\[(x+2)^2 + (y-1)^2 + (z-2)^2 = 5 \]

Step 7: Identify Center and Radius

The standard form of a sphere is \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \((h, k, l)\) is the center and \(r\) is the radius.For the equation \((x+2)^2 + (y-1)^2 + (z-2)^2 = 5\),- Center: \((-2, 1, 2)\)- Radius: \(\sqrt{5}\)

Key Concepts

Completing the Square

Completing the square is a mathematical process used to simplify quadratic equations and is often applied in geometry to find properties of figures like spheres. It's helpful in rewriting an equation into a form that easily shows the center and radius of a circle or sphere. Here's how you can do it:

• Identify the terms: Look for linear terms in the expression you want to transform. For example, in the expression \(x^2 + 4x\), \(4x\) is the linear term.
• Halve the coefficient: Take the coefficient of the linear term, divide it by 2, and square the result. For \(4x\), it's \((4/2)^2 = 4\).
• Add and subtract: Add and subtract this squared value within the expression to maintain balance. It turns \(x^2 + 4x\) into \((x + 2)^2 - 4\).

This method transforms equations to make them easier to solve and visualizes them in simpler geometric terms. Completing the square is routinely used for each variable component in the sphere equation formulation.

Center of a Sphere

The center of a sphere is a crucial geometric point that evenly spaces all outer points of the sphere. In the mathematical description of a sphere, the equation is often given in the form:
\[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]
This format reveals the center as the coordinate \((h, k, l)\). In the exercise you are solving, after completing the square for each term, the equivalent expression becomes:
\((x+2)^2 + (y-1)^2 + (z-2)^2 = 5\)
This allows us to directly identify the center of the sphere as \((-2, 1, 2)\). Every point from the center to the surface of the sphere is equidistant, which defines it perfectly in geometric space.

Radius of a Sphere

The radius of a sphere is the distance from the center to any point on the surface of the sphere. After completing the square and rearranging terms, the sphere equation \\((x + 2)^2 + (y - 1)^2 + (z - 2)^2 = 5\) is obtained.

Comparing this with the standard spherical equation \\((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\),
the expression on the right side, \(r^2 = 5\), shows that the radius \(r\) is actually \(\sqrt{5}\).

• The radius is always represented by the positive square root of the number on the right-hand side of the equation.
• It plays a major role in calculating the volume and surface area of a sphere.

To achieve a good understanding of a sphere's size and position, recognizing its center and radius is fundamental. This foundational knowledge assists in grasping more complex geometrical concepts involving planes and 3D shapes.