The line x + y = k meets the pair of straight lines $x^2+y^2-2x-4y+2=0$ in two points A and B. If O is the origin and $\angle AOB={90}^0$ then the value of k > 1 is 

$\begin{array}{l}GivenEquation{x}^2+y^2-2x-4y+2=0---\left(1\right)\\ andthelinex+y=k\\ \Rightarrow \frac{x+y}{k}=1--\left(2\right)\\ Homogenizing\left(1\right)with\left(2\right)\\ x^2+y^2-2x\left(\frac{x+y}{k}\right)-4y\left(\frac{x+y}{k}\right)+2{\left(\frac{x+y}{k}\right)}^2=0\\ \Rightarrow x^2+y^2-\frac{2x^2}{k}-\frac{2xy}{k}-\frac{4xy}{k}-\frac{4y^2}{k}+\frac{2x^2}{k^2}+\frac{2y^2}{k^2}+\frac{4xy}{k^2}=0\\ \because \angle AOB={90}^0\\ \Rightarrow x^{2}Co.eff+y^{2}Co.eff=0\\ \Rightarrow 1+1-\frac{2}{k}-\frac{4}{k}+\frac{2}{k^2}+\frac{2}{k^2}=0\\ \Rightarrow 2-\frac{6}{k}+\frac{4}{k^2}=0\\ \Rightarrow 2k^2-6k+4=0\\ \Rightarrow k^2-3k+2=0\\ \Rightarrow (k-1)(x-2)=0\\ \Rightarrow k=1,2[\because k>1]\\ \therefore k=2\end{array}$