因$$x
e 5$$，有$$\frac{\left\textbar{} 3-x \right\textbar}{\left\textbar{} 5-x \right\textbar}\leqslant \frac{1}{2}\Leftrightarrow 2\left\textbar{} 3-x \right\textbar\leqslant \left\textbar{} 5-x \right\textbar$$．  当$$x\leqslant 3$$时，$$2(3-x)\leqslant (5-x)$$，解得$$x\geqslant 1$$；  当$$3\textless{}x\textless{}5$$时，$$2(x-3)\leqslant (5-x)$$，解得$$x\leqslant \frac{11}{3}$$；  当$$x\textgreater5$$时，$$2(x-3)\leqslant x-5$$，无解．  所以$$M=\left { x\textbar1\leqslant x\leqslant \frac{11}{3} \right }$$．  又因$${{x}^{2}}-2x+c=0,x=1\pm \sqrt{1-c}$$，  所以$$N=\left { x\textbar1-\sqrt{1-c}\leqslant x\leqslant 1+\sqrt{1-c} \right }$$．  显然，$$1-\sqrt{1-c}\leqslant 1$$，  所以只要$$1+\sqrt{1-c}\geqslant \frac{11}{3}$$，即$$c\leqslant -\frac{55}{9}$$，即有$$M\cap N=M$$． 
若集合$$M=\left { x\left\textbar{} \frac{\left\textbar{} 3-x \right\textbar}{\left\textbar{} 5-x \right\textbar}\leqslant \frac{1}{2} \right. \right }$$和集合$$N= {x\textbar{{x}^{2}}-2x+c\leqslant 0 }$$满足$$M\cap N=M$$，则实数$$c$$的取值范围是．
Options:
A. $$c\leqslant -\frac{44}{9}$$ 
B. $$c\leqslant -\frac{55}{9}$$ 
C. $$c\leqslant -\frac{66}{9}$$ 
D. $$c\leqslant -\frac{77}{9}$$ 