Geometric solution? Find the locus of the point $M$ lie on the sphere so that $MA-2MB$ smallest

Question

Let be given the sphere has the equation $x^2 + y^2 + (z-3)^2 = 8$ and two points $A(4,4,3)$, $B(1,1,1)$. Find the locus of the points $M$ lie on the sphere so that $|MA-2MB|$ is smallest.

I tried. Put

$x=2\sqrt{2}\cos a \cdot \sin b$, $x=2\sqrt{2}\sin a \cdot \sin b$, $z=3+2\sqrt{2} \cos b.$

But, It's difficult for me to calculate.

Answer 1

It can be $0$.

Indeed, for which should be $$4((x-1)^2+(y-1)^2+(z-1)^2))=(x-4)^2+(y-4)^2+(z-3)^2,$$ where $M(x,y,z)$ or $$3(x^2+y^2+z^2)-2z=29$$ and since $x^2+y^2=8-(z-3)^2,$ we obtain $$3(8-(z-3)^2+z^2)-2z=29$$ or $$z=2,$$ which gives $$x^2+y^2=7.$$ Id est, we got a circle.

Answer 2

In the plane, the locus of points $M$ such that $MA=2MB$ is a circle, whose center $C$ lies on line $AB$ with $A<B<C$ and $BC={1\over3}AB$, and whose radius is ${2\over3}AB$.

In 3D space the locus is obviously a sphere with the same center and radius. In your case that sphere intersects the given sphere and this intersection is then the required locus.