The maximum value of  ${(\sqrt{-3+4x-x^2}+4)}^2+(x-5{)}^2.$ $\text{(where }1\le x\le 3\text{)}$ is

Here, ${(\sqrt{-3+4x-x^2}+4)}^2+(x-5{)}^2$, represents the square of the distance between circle

$y=\sqrt{-3+4x-x^2}$ and point (5 ,–4 ) ie, Maximum distance between $x^2+y^2-4x+3=0$  and (5,–4) aquared $\Rightarrow P{Q}^2=(\mathrm{PC}+\text{radius})={(\sqrt{(5-2{)}^2+(4-0{)}^2}+1)}^2$

$=6^2=36.$ Hence , (b) is the correct answer.