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(A) Given,the line is $x+y-\lambda=0$,so $\frac{x+y}{\lambda}=1$ ... $(i)$.Given the pair of lines $x^2+y^2-2x-4y+2=0$.To find the equation of the lin

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$2$

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(A) Given,the line is $x+y-\lambda=0$,so $\frac{x+y}{\lambda}=1$ ... $(i)$.Given the pair of lines $x^2+y^2-2x-4y+2=0$.To find the equation of the lines joining the origin to the points of intersection,we homogenize the equation of the curve using $(i)$:$x^2+y^2-2x(1)-4y(1)+2(1)^2=0$$x^2+y^2-2x(\frac{x+y}{\lambda})-4y(\frac{x+y}{\lambda})+2(\frac{x+y}{\lambda})^2=0$Multiplying by $\lambda^2$:$\lambda^2(x^2+y^2)-2x\lambda(x+y)-4y\lambda(x+y)+2(x+y)^2=0$$\lambda^2x^2+\lambda^2y^2-2x^2\lambda-2xy\lambda-4xy\lambda-4y^2\lambda+2x^2+2y^2+4xy=0$$(\lambda^2-2\lambda+2)x^2+(4-6\lambda)xy+(\lambda^2-4\lambda+2)y^2=0$Since $\angle AOB=90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(\lambda^2-2\lambda+2)+(\lambda^2-4\lambda+2)=0$$2\lambda^2-6\lambda+4=0$$\lambda^2-3\lambda+2=0$$(\lambda-1)(\lambda-2)=0$Thus,$\lambda=1$ or $\lambda=2$. Since $2$ is an option,the correct choice is $A$.

Suppose $A$ and $B$ are the points at which the line $x+y-\lambda=0$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$. If $\angle AOB=90^{\circ}$,then a value of $\lambda$ is

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