Chapter 10, Problem 18RE

To determine

To find: The center and radius of the sphere $x^2+y^2+z^2-10x+6y-4z+34=0$ .

Answer to Problem 18RE

The center and radius of the sphere $x^2+y^2+z^2-10x+6y-4z+34=0$ are $\left(5,-3,2\right)$ and $2$ , respectively.

Explanation of Solution

Given information:

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

Calculation:

The general standard form of the equation of sphere is ${\left(x-h\right)}^2+{\left(y-k\right)}^2+{\left(z-j\right)}^2=r^2$ with center $\left(h,k,j\right)$ and $r$ radius.

Simplify the given equation of sphere.

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

Compare the equation of sphere ${\left(x-5\right)}^2+{\left(y+3\right)}^2+{\left(z-2\right)}^2=2^2$ with general equation. It gives the value of $h$ , $k$ , $j$ as $5$ , $-3$ , $2$ respectively and the value of $r$ as $2$ .

Therefore, the center and radius of the sphere $x^2+y^2+z^2-4x-6y+2z+6=0$ are $\left(5,-3,2\right)$ and $2$ , respectively.