Chapter 9.3, Problem 24E

To determine

To find:The center and radius of sphere using distance formula.

Answer to Problem 24E

The standard form of the equation is ${\left(x-0\right)}^2+{\left(y+1\right)}^2+z^2=4$ with center $\left(0,-10\right)$ with radius $r=2$ .

Explanation of Solution

Given information:

The givensets of all points $\left(x,y,z\right)$ in a coordinate space that are twice as far from the points $Q\left(0,3,0,\right)$ , as far from the point $P\left(0,0,0,\right)$ .

Calculation:

The distance between the points $A\left(x,y,z\right)$ and origin.

$\begin{array}{l}{d}_1=\sqrt{{\left(x-0\right)}^2+{\left(y-0\right)}^2+{\left(z-0\right)}^2}\\ d_1=\sqrt{x^2+y^2+z^2}\end{array}$

Calculate the distance between the points $R\left(x,y,z\right)$ and $\left(0,3,0,\right)$ .

$\begin{array}{l}{d}_2=\sqrt{{\left(x-0\right)}^2+{\left(y-3\right)}^2+{\left(z-0\right)}^2}\\ d_2=\sqrt{x^2+{\left(y-3\right)}^2+z^2}\end{array}$

Using the given information $d_2=2d_1$ .

$\sqrt{x^2+{\left(y-3\right)}^2+z^2}=\sqrt{x^2+y^2+z^2}$

Squaring on both sides of the above equation.

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

  ${\left(x-0\right)}^2+{\left(y+1\right)}^2+z^2=4$

The standard form of the equation is ${\left(x-0\right)}^2+{\left(y+1\right)}^2+z^2=4$ with center $\left(0,-10\right)$ with radius $r=2$ .

Therefore, thestandard form of the equation is ${\left(x-0\right)}^2+{\left(y+1\right)}^2+z^2=4$ with center $\left(0,-10\right)$ with radius $r=2$ .