jmc

Let $z=x+yi,$ where $x$ and $y$ are real numbers. Then $|x+yi-5-i|=|(x-5)+(y-1)i|=5,$ so $$(x-5{)}^2+(y-1{)}^2=25.$$This simplifies to $x^2-10x+y^2-2y=-1.$

Also, $$\begin{array}{rl}|z-1+2i{|}^2+|z-9-4i{|}^2& =|x+yi-1+2i{|}^2+|x+yi-9-4i{|}^2\\ & =|(x-1)+(y+2)i{|}^2+|(x-9)+(y-4)i{|}^2\\ & =(x-1{)}^2+(y+2{)}^2+(x-9{)}^2+(y-4{)}^2\\ & =2x^2-20x+2y^2-4y+102\\ & =2(x^2-10x+y^2-2y)+102\\ & =2(-1)+102=100.\end{array}$$Thus, the expression is always equal to $\boxed{100}.$

Geometrically, the condition $|z-5-i|=5$ states that $z$ lies on a circle centered at $5+i$ with radius 5.



Note that $1-2i$ and $9+4i$ are diametrically opposite on this circle. Hence, when we join $z$ to $1-2i$ and $9+4i,$ we obtain a right angle. Thus, the expression in the problem is equal to the square of the diameter, which is $10^2=100.$