Chapter 11.1, Problem 59E

To determine

Tofind:Thecenter and radius of the sphere when the equation of the sphere is $9x^2+9y^2+9z^2-18x-6y-72z+73=0$ .

Answer to Problem 59E

Thecenter and radius of the sphere is $\left(1,\frac{1}{3},4\right)$ and3, respectively.

Explanation of Solution

Given information:

The given equation is $9x^2+9y^2+9z^2-18x-6y-72z+73=0$ .

Calculation:

Represent the given equation in the standard form of the equation of the sphere.

  $\begin{array}{c}9x^2+9y^2+9z^2-18x-6y-72z+73=0\\ x^2+y^2+z^2-2x-\frac{6}{9}y-8z+\frac{73}{9}=0\left\{\text{Take}\text{9}\text{common}\text{on}\text{both}\text{the}\text{sides}\right\}\\ \left(x^2-2x+1\right)+\left(y^2-\frac{2}{3}y+\frac{1}{9}\right)+\left(z^2-8z+16\right)=1+16+\frac{1}{9}-\frac{73}{9}\\ {\left(x-1\right)}^2+{\left(y-\frac{1}{3}\right)}^2+{\left(z-4\right)}^2=17-\frac{72}{9}\\ {\left(x-1\right)}^2+{\left(y-\frac{1}{3}\right)}^2+{\left(z-4\right)}^2=3^2\end{array}$

Compare the above equation with the standard form of the equation of the sphere where $\left(h,k,j\right)$ is the center and the radius is $r$ .

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

From the above, the center of the sphere is $\left(h,k,j\right)=\left(1,\frac{1}{3},4\right)$ and the radius is $r=3$ .

Therefore, thecenter and radius of the sphere is $\left(1,\frac{1}{3},4\right)$ and 3, respectively.