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The line $x+y=k$ meets the pair of straight lines $x^2+y^2-2 x-4 y+2=0$ at two points $A$ and $B$. If $O$ is the origin and $\angle A O B=90^{\circ}$, then the value of $k(>1)$ is
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The correct answer is:
2
Homogenise the pair of lines $x^2+y^2-2 x-4 y+2=0$ through $x+y=k$, we get
$$
\begin{aligned}
& x^2+y^2-2 x\left(\frac{x+y}{k}\right)-4 y\left(\frac{x+y}{k}\right) \\
&+2\left(\frac{x+y}{k}\right)^2=0
\end{aligned}
$$
Since, intersection points of line and pair of lines make an angle $90^{\circ}$ at origin $O$.
$\therefore$ Coefficient of $x^2+$ Coefficient of $y^2=0$
$$
\begin{aligned}
& \Rightarrow\left(1-\frac{2}{k}+\frac{2}{k^2}\right)+\left(1-\frac{4}{k}+\frac{2}{k^2}\right)=0 \\
& \Rightarrow \quad 2-\frac{6}{k}+\frac{4}{k^2}=0 \\
& \Rightarrow \quad k^2-3 k+2=0 \\
& \Rightarrow \quad(k-2)(k-1)=0 \\
& \Rightarrow \quad k=1,2 \\
& \text { But } k>1 \\
& \therefore \quad k=2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& x^2+y^2-2 x\left(\frac{x+y}{k}\right)-4 y\left(\frac{x+y}{k}\right) \\
&+2\left(\frac{x+y}{k}\right)^2=0
\end{aligned}
$$
Since, intersection points of line and pair of lines make an angle $90^{\circ}$ at origin $O$.
$\therefore$ Coefficient of $x^2+$ Coefficient of $y^2=0$
$$
\begin{aligned}
& \Rightarrow\left(1-\frac{2}{k}+\frac{2}{k^2}\right)+\left(1-\frac{4}{k}+\frac{2}{k^2}\right)=0 \\
& \Rightarrow \quad 2-\frac{6}{k}+\frac{4}{k^2}=0 \\
& \Rightarrow \quad k^2-3 k+2=0 \\
& \Rightarrow \quad(k-2)(k-1)=0 \\
& \Rightarrow \quad k=1,2 \\
& \text { But } k>1 \\
& \therefore \quad k=2 \\
&
\end{aligned}
$$
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