Chapter 10.1, Problem 73E

To determine

To find: The center and radius of the sphere $4x^2+4y^2+4z^2-8x+16y-1=0$ .

Answer to Problem 73E

The center and radius of the sphere $4x^2+4y^2+4z^2-8x+16y-1=0$ are $\left(1,-2,0\right)$ and $\frac{\sqrt{21}}{2}$ respectively.

Explanation of Solution

Given information:

The equation of the sphere is $4x^2+4y^2+4z^2-8x+16y-1=0$ .

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.

  $\begin{array}{c}4x^2+4y^2+4z^2-8x+16y-1=0\\ 4\left(x^2-2x\right)+4\left(y^2+4y\right)+4z^2=1\\ 4\left(x^2-2x+1^2-1^2\right)+4\left(y^2+4y+2^2-2^2\right)+4z^2=1\\ 4{\left(x-1\right)}^2+4{\left(y+2\right)}^2+4z^2-4-16=1\end{array}$

Further simplify,

  $\begin{array}{c}4{\left(x-1\right)}^2+4{\left(y+2\right)}^2+4z^2-4-16=1\\ 4{\left(x-1\right)}^2+4{\left(y+2\right)}^2+4z^2=21\\ {\left(x-1\right)}^2+{\left(y+2\right)}^2+z^2=\frac{21}{4}\\ {\left(x-1\right)}^2+{\left(y+2\right)}^2+z^2={\left(\frac{\sqrt{21}}{2}\right)}^2\end{array}$

Compare the equation of sphere ${\left(x-1\right)}^2+{\left(y+2\right)}^2+z^2={\left(\frac{\sqrt{21}}{2}\right)}^2$ with general equation. It gives the value of $h$ , $k$ , $j$ as $1$ , $-2$ , $0$ respectively and the value of $r$ as $\frac{\sqrt{21}}{2}$ .

Therefore, the center and radius of the sphere $4x^2+4y^2+4z^2-8x+16y-1=0$ are $\left(1,-2,0\right)$ and $\frac{\sqrt{21}}{2}$ respectively.