Chapter 11, Problem 23RE

To determine

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

Answer to Problem 23RE

The center is $\left(5,–3,4\right)$ and radius is 4

Explanation of Solution

Given:

The equation of the sphere $x^2+y^2+z^2–10x+6y–4z+34=0$

Consider the equation of the sphere $x^2+y^2+z^2=r^2$

The center and radius can be obtained by completing the square in the equation of sphere.

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

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

Consider the equation of the sphere $x^2+y^2+z^2–10x+6y–4z+34=0$

To obtain the standard equation of the sphere, complete the square as follows:

  ${\begin{array}{l}\left(x^2–10x\right)+\left(y^2+6y\right)+\left(z^2–4z\right)+34=0\hfill \\ \left(x^2–10x+25\right)+\left(y^2+6y+9\right)+\left(z^2–4z+16\right)+34–25–9–16=0\hfill \\ {\left(x–5\right)}^2+{\left(y+3\right)}^2+\left(z–4\right)2=16\hfill \\ {\left(x–5\right)}^2+{\left(y+3\right)}^2+\left(z–4\right)2={\left(4\right)}^2\hfill \end{array}}^{}$

Compare the above equation by the general equation of sphere that is

  ${(x-h)}^2+{(y-k)}^2+{(z-j)}^2={(r)}^2$ , the radius and center is obtained.

Hence, the center is $\left(5,–3,4\right)$ and radius is 4.