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(D) Homogenize the equation of the pair of lines $x^2+y^2-2x-4y+2=0$ using the line $x+y=k$,which can be written as $\frac{x+y}{k}=1$. Substituting th

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$2$

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(D) Homogenize the equation of the pair of lines $x^2+y^2-2x-4y+2=0$ using the line $x+y=k$,which can be written as $\frac{x+y}{k}=1$. Substituting this into the equation,we get: $x^2+y^2-2x(\frac{x+y}{k})-4y(\frac{x+y}{k})+2(\frac{x+y}{k})^2=0$. Since the lines $OA$ and $OB$ are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero. Expanding the equation: $x^2+y^2-\frac{2}{k}(x^2+xy)-\frac{4}{k}(xy+y^2)+\frac{2}{k^2}(x^2+2xy+y^2)=0$. Grouping the terms: $(1-\frac{2}{k}+\frac{2}{k^2})x^2 + (1-\frac{4}{k}+\frac{2}{k^2})y^2 + (\dots)xy = 0$. Setting the sum of coefficients of $x^2$ and $y^2$ to zero: $(1-\frac{2}{k}+\frac{2}{k^2}) + (1-\frac{4}{k}+\frac{2}{k^2}) = 0$. $2 - \frac{6}{k} + \frac{4}{k^2} = 0$. Multiplying by $k^2$: $2k^2 - 6k + 4 = 0 \Rightarrow k^2 - 3k + 2 = 0$. $(k-1)(k-2) = 0$. Thus,$k=1$ or $k=2$. Given $k>1$,the value is $k=2$.

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

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