Chapter 11.1, Problem 60E

To determine

Tofind:Thestandard equation of the sphere with the endpoints of the diameter as $\left(1,0,0\right)$ and $\left(0,5,0\right)$ .

Answer to Problem 60E

Thestandard equation of the sphere with the endpoints of the diameter as $\left(1,0,0\right)$ and $\left(0,5,0\right)$ is ${\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{5}{2}\right)}^2+{\left(z\right)}^2=\frac{26}{4}$ .

Explanation of Solution

Given information:

The endpoints of the diameter are $\left(1,0,0\right)$ and $\left(0,5,0\right)$ .

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.

  $\begin{array}{c}\text{Midpoint}=\left(\frac{1+0}{2},\frac{0+5}{2},\frac{0+0}{2}\right)\\ =\left(\frac{1}{2},\frac{5}{2},0\right)\end{array}$

So, the center of the sphereis $\left(\frac{1}{2},\frac{5}{2},0\right)$ .

Calculate the radius of the sphere when the center is $\left(\frac{1}{2},\frac{5}{2},0\right)$ and the endpoint is $\left(1,0,0\right)$ .

  $\begin{array}{c}\text{Radius}=\sqrt{{\left(1-\frac{1}{2}\right)}^2+{\left(0-\frac{5}{2}\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(-\frac{5}{2}\right)}^2}\\ =\sqrt{\frac{1}{4}+\frac{25}{4}}\\ =\sqrt{\frac{26}{4}}\text{units}\end{array}$

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 $\left(\frac{1}{2},\frac{5}{2},0\right)$ and $\sqrt{\frac{26}{4}}$ for $r$ in the above equation to calculate the standard equation of the sphere.

  $\begin{array}{c}{\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{5}{2}\right)}^2+{\left(z-0\right)}^2={\left(\sqrt{\frac{26}{4}}\right)}^2\\ {\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{5}{2}\right)}^2+{\left(z\right)}^2=\frac{26}{4}\end{array}$

Therefore, the standard equation of the sphere with the endpoints of the diameter as $\left(1,0,0\right)$ and $\left(0,5,0\right)$ is ${\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{5}{2}\right)}^2+{\left(z\right)}^2=\frac{26}{4}$ .