Given a complex number $z$ that $|z-2-i|=2\sqrt{2}$. Find the maximum and minimum of $H=|z+3-2i|+|z-3+4i|$.
The minimum is easy to find: $H=|z+3-2i|+|z-3+4i| \geq|z+3-2i-z+3-4i|=6\sqrt{2}$
I am struggling with the maximum. It leads to find the maximum of $\sqrt{(a+3)^2+(b-2)^2}+\sqrt{(a-3)^2+(b+4)^2}$ given $(a-2)^2+(b-1)^2=8$.
I have drawn these on the coordinate plane. Let $A(-3,2), B(3,-4)$, we need to find pint $M$ on circle $(x-2)^2+(y-1)^2=8$ such that $MA+MB$ attains its minimum or maximum.
Then I percieve that when M is located at $C$ and $E$, $MA+MB$ attains its minimum and maximum respectively. (surprisingly, $C$ is the tangent point).
But I can't give the rigorous proof of that either analytically of geometrically. Hope your kind help!
This question involves a circle
$$ C\rightarrow |z-2+i| = 2\sqrt2 $$
and a generic ellipse $E$ with focus at $z_1 = -3+2i$ and $z_2 = 3-4i$
The problem can be asked as:
Determine the ellipse with focus at $z_1,z_2$ tangent to the cylinder $C$
The problem generally has two solutions: one minimal and other maximal. In this case only the maximal. The minimum is the distance $|z_1-z_2|$ (a segment as a degenerated ellipse).