Chapter 11.1, Problem 59E

To determine

Tofind:Thestandard equation of the sphere with the endpoints of the diameter as $\left(3,0,0\right)$ and $\left(0,0,6\right)$ .

Answer to Problem 59E

Thestandard equation of the sphere with the endpoints of the diameter as $\left(3,0,0\right)$ and $\left(0,0,6\right)$ is ${\left(x-\frac{3}{2}\right)}^2+{\left(y\right)}^2+{\left(z-3\right)}^2=\frac{45}{4}$ .

Explanation of Solution

Given information:

The endpoints of the diameter are $\left(3,0,0\right)$ and $\left(0,0,6\right)$ .

Calculation:

The endpoints of the diameter lies on the sphere. The midpoint between these two endpoints will be the center of the sphere.

Calculate the midpoint of the sphere.

  $\begin{array}{c}\text{Midpoint}=\left(\frac{3+0}{2},\frac{0+0}{2},\frac{0+6}{2}\right)\\ =\left(\frac{3}{2},0,3\right)\end{array}$

So, the center of the sphereis $\left(\frac{3}{2},0,3\right)$ .

Calculate the radius of the sphere when the center is $\left(\frac{3}{2},0,3\right)$ and the endpoint is $\left(3,0,0\right)$ .

  $\begin{array}{c}\text{Radius}=\sqrt{{\left(3-\frac{3}{2}\right)}^2+{\left(0-0\right)}^2+{\left(0-3\right)}^2}\\ =\sqrt{{\left(\frac{3}{2}\right)}^2+0+{\left(-3\right)}^2}\\ =\sqrt{\frac{9}{4}+9}\\ =\sqrt{\frac{45}{4}}\text{units}\end{array}$

The standard equation of a sphere with center $\left(h,k,j\right)$ and radius $r$ can be calculated as follows.

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

Substitute $\left(\frac{3}{2},0,3\right)$ and $\sqrt{\frac{45}{4}}$ for $r$ in the above equation to calculate the standard equation of the sphere.

  $\begin{array}{c}{\left(x-\frac{3}{2}\right)}^2+{\left(y-0\right)}^2+{\left(z-3\right)}^2={\left(\sqrt{\frac{45}{4}}\right)}^2\\ {\left(x-\frac{3}{2}\right)}^2+{\left(y\right)}^2+{\left(z-3\right)}^2=\frac{45}{4}\end{array}$

Therefore, the standard equation of the sphere with the endpoints of the diameter as $\left(3,0,0\right)$ and $\left(0,0,6\right)$ is ${\left(x-\frac{3}{2}\right)}^2+{\left(y\right)}^2+{\left(z-3\right)}^2=\frac{45}{4}$ .