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Chapter 13: Problem 44

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

Short Answer

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

Step by step solution

01

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鈥檚 equation. The job now is to rearrange the equation and complete the squares for x, y, and z.
02

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:\[x^{2} + (y^{2} - 4y) + (z^{2} + 6z) + 4=0\]
03

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:\[x^{2}+(y-2)^{2}+(z+3)^{2}-9=0\]
04

Express in standard formula

Rearranging the equation gives us:\[(x-0)^{2}+(y-2)^{2}+(z+3)^{2}=9\]
05

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.

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