jmc

From the equation $x+\frac{1}{x}=5,$ $x^2+1=5x,$ so $$x^2=5x-1.$$Then $$(x-2{)}^2=x^2-4x+4=(5x-1)-4x+4=x+3.$$Hence, $$\begin{array}{rl}(x-2{)}^2+\frac{25}{(x-2{)}^2}& =x+3+\frac{25}{x+3}\\ & =\frac{(x+3{)}^2+25}{x+3}\\ & =\frac{x^2+6x+9+25}{x+3}\\ & =\frac{(5x-1)+6x+34}{x+3}\\ & =\frac{11x+33}{x+3}\\ & =\boxed{11}.\end{array}$$