The range of values of r for which the point $\left(-5+\frac{\mathrm{r}}{\sqrt{2}},-3+\frac{\mathrm{r}}{\sqrt{2}}\right)$is an interior point of the major segment of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=16$, cut-off by the line $\mathrm{x}+\mathrm{y}=2,$is 

The given point is an interior point. Therefore, 
$$\begin{gathered} {\left(-5+\frac{\mathrm{r}}{\sqrt{2}}\right)}^2+{\left(-3+\frac{\mathrm{r}}{\sqrt{2}}\right)}^2-16<0 \\ \mathrm{or} {\mathrm{r}}^2-8\sqrt{2}\mathrm{r}+10<0 \\ \mathrm{or} 4\sqrt{2}-\sqrt{14}<\mathrm{r}<4\sqrt{2}+\sqrt{14} \left(\mathrm{i}\right) \end{gathered}$$
The point is on the major segment., Therefore, the center and the point are on the same side of the line x + y = 2. Thus, 
$$\begin{gathered} -5+\frac{\mathrm{r}}{\sqrt{2}}-3+\frac{\mathrm{r}}{\sqrt{2}}-2<0 \\ \mathrm{or} \mathrm{r}<5\sqrt{2} \left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), 4\sqrt{2}-\sqrt{14}<\mathrm{r}<5\sqrt{2}. \end{gathered}$$