Let maximum value of the expression y=|k-3| 2 x+|t-4| 2 x+3 where 0...

\begin{aligned} \left.y\right|_{\max .} & =\sqrt{(k-3)^2+(t-4)^2}+3=6 \\ (k-3)^2+(t-4)^2 & =9 \end{aligned} Put $$k-3=3 \cos \theta$$ Put $$t+4=3 \sin \theta$$\begin{aligned} k^2+t^2 & =(3+3 \cos \theta)^2+(-4+3 \sin \theta)^2 \\ & =34+18 \cos \theta-24 \sin \theta \\ +\left.t^2\right|_{\min .} & =(34-30)=4 \end{aligned}$$\therefore \quad k^2+\left.t^2\right|_{\min .}=(34-30)=4$$