Problem 15 - Algebra and Trigonometry

Question

\(15-18=\) Show that the equation represents a sphere, and find its center and radius. $$ x^{2}+y^{2}+z^{2}-10 x+2 y+8 z=9 $$

Step-by-Step Solution

Answer

The sphere's center is (5, -1, -4) and its radius is \(\sqrt{51}\).

Step 1: Rewrite the equation in standard form

To find the center and radius of a sphere given by an equation, we need to rewrite the equation in the standard form of a sphere, which is: \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). To do this, we complete the square for each variable. Start by grouping the terms involving each variable: \(x^2 - 10x\), \(y^2 + 2y\), and \(z^2 + 8z\).

Step 2: Complete the square for x-terms

For the terms \(x^2 - 10x\), complete the square. Add and subtract \((\frac{-10}{2})^2 = 25\). The expression becomes \((x-5)^2 - 25\).

Step 3: Complete the square for y-terms

For the terms \(y^2 + 2y\), complete the square. Add and subtract \((\frac{2}{2})^2 = 1\). The expression becomes \((y+1)^2 - 1\).

Step 4: Complete the square for z-terms

For the terms \(z^2 + 8z\), complete the square. Add and subtract \((\frac{8}{2})^2 = 16\). The expression becomes \((z+4)^2 - 16\).

Step 5: Simplify and finalize the sphere equation

Substitute the completed squares back into the original equation: \((x-5)^2 - 25 + (y+1)^2 - 1 + (z+4)^2 - 16 = 9\). Simplify it to \((x-5)^2 + (y+1)^2 + (z+4)^2 = 9 + 25 + 1 + 16\). This reduces to \((x-5)^2 + (y+1)^2 + (z+4)^2 = 51\).

Step 6: Identify the center and radius

The equation \((x-5)^2 + (y+1)^2 + (z+4)^2 = 51\) is in the standard form of a sphere. Thus, the center of the sphere is \((5, -1, -4)\) and the radius is \(\sqrt{51}\).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. For a variable with the form of \( ax^2 + bx \), this involves making it into the form \((x - h)^2\) by adding and subtracting a specific number.

• First, take the coefficient of the linear term \( b \) (which is the term without any exponent)
• Divide it by 2, and square the result.
• Add this new value inside the parentheses to complete the square.
• Don't forget to subtract the same value outside the parentheses to keep the equation balanced.

For example, we consider the expression \( x^2 - 10x \) from our problem. By taking half of \(-10\), which results in \(-5\), and squaring it, we get 25. Therefore, the expression can be rewritten as \((x - 5)^2 - 25\). This technique is essential because it allows us to express the equation of a sphere in its standard form, making it easy to find both the center and the radius.

Center of Sphere

The center of a sphere is a fixed point at an equal distance from all points on the surface of the sphere. In the standard form of the sphere equation, \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), the variables \( h, k, \) and \( l \) represent the coordinates of the center of the sphere: \((h, k, l)\).

• Here, \( x-h \) indicates a shift of \( h \) units in the x-direction
• \( y-k \) shows a shift of \( k \) units in the y-direction
• \( z-l \) represents a shift of \( l \) units in the z-direction

These shifts determine how the sphere is positioned in 3D space. In our solution, by completing the square for each group of terms and setting them to match the standard form, we find the equation to be \((x-5)^2 + (y+1)^2 + (z+4)^2 = 51\). Thus, the center of this sphere is determined to be at point \((5, -1, -4)\). This tells us exactly how much to translate the sphere from the origin to its correct location.

Radius of Sphere

The radius of a sphere is the distance from the center to any point on the surface of the sphere. It is a fixed length that defines the size of the sphere. Once the equation of the sphere is in the form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), the term on the right, \( r^2 \), represents the square of the radius.

• To find the actual radius, you take the square root of \( r^2 \).
• This means if your equation is \( (x-5)^2 + (y+1)^2 + (z+4)^2 = 51 \), then the radius \( r \) is \( \sqrt{51} \).

Calculating the radius directly provides a measure of how large or small the sphere is. In our problem, \( \sqrt{51} \) gives us an exact measurement of the sphere's radius, ensuring all points on the sphere's surface are precisely this distance from the center \((5, -1, -4)\). This property helps in visualizing and understanding the sphere's dimension and position in space.