Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Find the sphere's center and radius. $$ x^{2}+y^{2}+z^{2}-4 y+6 z+4=0 $$

Step-by-Step Solution

Answer

The sphere's center is at (0,2,-3) and its radius is 3.

Step 1: Identify the components of the equation

Look at each component in the given equation. The terms \(x^{2}\), \(y^{2}\) and \(z^{2}\) match our generic form of a sphere’s equation. The job now is to rearrange the equation and complete the squares for x, y, and z.

Step 2: Group the like terms

Rewrite the equation grouping the like terms together. This regroups the equation into three squares, making it easier to identify the sphere's center and radius. \[x^{2} + y^{2} - 4y + z^{2} + 6z + 4=0 \]This can be rearranged as:\[x^{2} + (y^{2} - 4y) + (z^{2} + 6z) + 4=0\]

Step 3: Complete the squares

To make the equation look like our standard form, we need to complete the square for y and z terms. In general, for a term like \(y^2 - 2ay\), we could convert it to \((y-a)^2\) by adding \(a^2\).So, for \(y^2 - 4y\), we add and subtract \((-4/2)^2=4\). And for \(z^2 + 6z\), we add and subtract \((6/2)^2=9\).This leads us to:\[x^{2} +(y^{2} - 4y + 4) + (z^{2} + 6z +9) - 4 - 9 + 4 = 0 \]Which simplifies to:\[x^{2}+(y-2)^{2}+(z+3)^{2}-9=0\]

Step 4: Express in standard formula

Rearranging the equation gives us:\[(x-0)^{2}+(y-2)^{2}+(z+3)^{2}=9\]

Step 5: Identify the center and the radius

This equation is in the standard form of a sphere's equation. Comparing this equation with the standard equation, we find that the center of the sphere is at (0,2,-3 ) and the radius of the sphere is \(\sqrt{9}\), which is 3.