Chapter 11, Problem 19RE

To determine

Find the standard form of the equation of the sphere and radius.

Answer to Problem 19RE

The standard form of the equation of the sphere whose center $(0,0,0)$ and radius $2\sqrt{3}$ is $x^2+y^2+z^2=12$

Explanation of Solution

Given:

The end points of the diameter

  $\left(–2,–2,–2\right)and\left(2,2,2\right)$

Consider the following end points of the diameter

  $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$

The center of the sphere is mid-point of the diameter and therefore, the mid-point of the diameter is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

The distance between the points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ in space is

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

The standards form of the equation of the sphere whose center $(h,k,j)$ and radius r is

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

  $\text{Radius}=\frac{1}{2}(\text{diameter})$

Consider the following end points of the diameter

  $\left(–2,–2,–2\right)and\left(2,2,2\right)$

Since the center of the sphere is mid-point of the diameter and therefore, the mid-point of the diameter is

  $\begin{array}{cc}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)& =\left(\frac{-2+2}{2},\frac{-2+2}{2},\frac{-2+2}{2}\right)\\ & =\left(\frac{0}{2},\frac{0}{2},\frac{0}{2}\right)\\ & =(0,0,0)\end{array}$

Hence the center is $(0,0,0)$

Use the distance formula in space d= NA to find the distance between two points (−2,−2,−2) and (2,2,2)

Therefore,

  $\begin{array}{cc}d& =\sqrt{{(2+2)}^2+{(2+2)}^2+{(2+2)}^2}\\ & =\sqrt{{(4)}^2+{(4)}^2+{(4)}^2}\end{array}$

Simplify,

  $\begin{array}{l}=\sqrt{16+16+16}\hfill \\ =\sqrt{48}\hfill \\ =4\sqrt{3}\hfill \end{array}$

Therefore, the radius of sphere is

  $\begin{array}{cc}r& =\frac{1}{2}(4\sqrt{3})\\ & =2\sqrt{3}\end{array}$

Use standard form of the equation of the sphere is ${(x-h)}^2+{(y-k)}^2+{(z-j)}^2={(r)}^2$

Therefore,

  $\begin{array}{cc}{(x-0)}^2+{(y-0)}^2+{(z-0)}^2& ={(2\sqrt{3})}^2\\ x^2+y^2+z^2& =12\end{array}$

Therefore, standard form of the equation of the sphere whose center $(0,0,0)$ and radius $2\sqrt{3}$ is $x^2+y^2+z^2=12$