Chapter 11, Problem 22RE

To determine

Find the center and radius of the sphere of given equation of sphere.

Answer to Problem 22RE

The center is $\left(2,3,0\right)$ and radius is 3.

Explanation of Solution

Given:

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

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 $x^2+y^2+z^2–4x–6y+4=0$

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

  ${\begin{array}{l}\left(x^2–4x\right)+\left(y^2–6y\right)+z^2+4=0\hfill \\ \left(x^2–4x+4\right)+\left(y^2–6y+9\right)+z^2+4–4–9=0\hfill \\ {\left(x–2\right)}^2+{\left(y–3\right)}^2+z2=9\hfill \\ {\left(x–2\right)}^2+{\left(y–3\right)}^2+z^2={\left(3\right)}^2\hfill \end{array}}^{}$

By comparing 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 $\left(2,3,0\right)$ and radius is 3.