Chapter 11, Problem 6CLT

To determine

To calculate:

The standard form of the equation of the sphere for the following points

  $\begin{array}{l}\left(0,0,0\right)\\ \left(4,4,8\right)\end{array}$

Answer to Problem 6CLT

The standard forms of the equation of the sphere whose radius $r=2\sqrt{6}$ and center $C=\left(2,2,4\right)$ is ${\left(x-2\right)}^2+{\left(y-2\right)}^2+{\left(z-4\right)}^2=24$ .

Explanation of Solution

Given information:

Endpoint of the diameter:

  $\begin{array}{l}A=\left(0,0,0\right)\\ B=\left(4,4,8\right)\end{array}$

Formula Used:

By using the formula:

  $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2+{\left(z-j\right)}^2=r^2$

Calculation:

The midpoint of the line segments the points:

  $\begin{array}{l}A=\left(x_1,y_1,z_1\right)\\ B=\left(x_2,y_2,z_2\right)\end{array}$

By using the formula:

  $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

The sphere center lies on the midpoint of diameter:

  $\begin{array}{l}C=\left(\frac{0+4}{2},\frac{0+4}{2},\frac{0+8}{2}\right)\\ C=\left(\frac{4}{2},\frac{4}{2},\frac{8}{2}\right)\\ C=\left(2,2,4\right)\end{array}$

By using the formula of distance between two points:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

The distance between any endpoint of the diameter and center of the sphere is radius:

  $\begin{array}{l}r=\sqrt{{\left(0-2\right)}^2+{\left(0-2\right)}^2+{\left(0-4\right)}^2}\\ r=\sqrt{{\left(-2\right)}^2+{\left(-2\right)}^2+{\left(-4\right)}^2}\\ r=\sqrt{4+4+16}\\ r=\sqrt{24}\\ r=2\sqrt{6}\end{array}$

Equation of the sphere with center:

  $\left(h,k,j\right)$

By using the formula:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2+{\left(z-j\right)}^2=r^2$

Put the value in formula:

  $\begin{array}{l}{\left(x-2\right)}^2+{\left(y-2\right)}^2+{\left(z-4\right)}^2={\left(2\sqrt{6}\right)}^2\\ {\left(x-2\right)}^2+{\left(y-2\right)}^2+{\left(z-4\right)}^2=24\end{array}$

The equation of the sphere whose radius $r=2\sqrt{6}$ and center $C=\left(2,2,4\right)$ is ${\left(x-2\right)}^2+{\left(y-2\right)}^2+{\left(z-4\right)}^2=24$