1708
\hspace{-16}$If Line $\mathbf{y=kx+5}$ is tangent to the ciuve at point\\\\ $\mathbf{y=x^2-2x+7}\;,$ Then Slope of These two curves at point \\\\ $\mathbf{\left(x_{1},y_{1}\right)}$ are equal. \\\\ So $\mathbf{{2x-2}|_{x_{1},y_{1}}={k}|_{x_{1},y_{1}}}$\\\\ So $\mathbf{2x_{1}-2=k\Leftrightarrow x_{1}=\frac{k+2}{2}}$\\\\ And $\mathbf{\left(x_{1},y_{1}\right)}$ also lie on $\mathbf{y=kx+5}$ and $\mathbf{y=x^2-2x+7}$\\\\ So $\mathbf{x^2_{1}-2x_{1}+7=kx_{1}+5}$\\\\ $\mathbf{\left(\frac{k+2}{2}\right)^2-2.\left(\frac{k+2}{2}\right)+7=k.\left(\frac{k+2}{2}\right)+5}$\\\\ After Solving We Get\\\\ $\mathbf{\boxed{\bold{k=\left(-2-2.\sqrt{2}\right)}}}$\\\\
