An object's Position $P$ changes so that its distance from $(1,2,-3)$ is always twice its distance from $(1,2,3).$ Show $P$ is on a sphere. Find its center and radius.
If $A(1,2,-3)$ and $P (x,y,z),$ the distance $AP =\sqrt{(x-1)^2+(y-2)^2+(z+3)^2}.$
If $B(1,2,3)$ and $P(x,y,z),$ the distance $BP =\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}.$
So, $$\sqrt{(x-1)^2+(y-2)^2+(z+3)^2} = 2\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}$$ $$(x-1)^2+(y-2)^2+(z+3)^2 = 4\big[(x-1)^2+(y-2)^2+(z-3)^2\big]$$
I did until there but how can I have an equation of the sphere , center and radius?
From:
$$(x-1)^2+(y-2)^2+(z+3)^2 = 4[(x-1)^2+(y-2)^2+(z-3)^2]$$
we get $$ 3(x-1)^2+3(y-2)^2+3z^2-30z+27=0$$
so $$ (x-1)^2+(y-2)^2+(z-5)^2=16$$
So the center of this sphere is at $(1,2,4)$ and radius is $4$.