Answer
center $(0,-1,0)$, radius $2$
Work Step by Step
Step 1. Use the distance formula, to point Q: $d_1^2=x^2+(y-3)^2+z^2$, for point P: $d_2^2=x^2+y^2+z^2$
Step 2. Since $d_1=2d_2$, we have $d_1^2=4d_2^2$
Step 3. Combine the above equations, we get $d_1^2=x^2+(y-3)^2+z^2=4x^2+4y^2+4z^2$ which can be simplified to $3x^2+3y^2+6y+3z^2=9$ or $x^2+y^2+2y+z^2=3$. Form a perfect square for $y$, we get
$x^2+(y+1)^2+z^2=4$
Step 4. We conclude that the result is a sphere centered at $(0,-1,0)$ with radius $r=2$
