Suppose $A$ and $B$ are the points at which the line $x+y-\lambda=0$ meets the pair of straight lines…
Suppose $A$ and $B$ are the points at which the line $x+y-\lambda=0$ meets the pair of straight lines $x^2+y^2-2 x-4 y+2=0$. If $\angle A O B=90^{\circ}$, then a value of $\lambda$ is
$2$
$3$
$4$
$0$
Solution
Given, line is $x+y-\lambda=0$
$\Rightarrow \quad x+y=\lambda$
$\Rightarrow \quad \frac{x+y}{\lambda}=1$...(i)
and pair of straight lines is
$x^2+y^2-2 x-4 y+2=0$
We will make homogeneous of the above line.
$x^2+y^2-2 x(\mathrm{l})-4 y(\mathrm{l})+2(\mathrm{l})^2=0$
$\Rightarrow \quad x^2+y^2-2 x\left(\frac{x+y}{\lambda}\right)-4 y\left(\frac{x+y}{\lambda}\right)$
$+2\left(\frac{x+y}{\lambda}\right)^2=0 \quad$ [using Eq. (i)]
$\Rightarrow \lambda^2\left(x^2+y^2\right)-2 x^2 \lambda-2 y x \lambda-4 x y \lambda-4 y^2 \lambda$ $+2\left(x^2+y^2+2 x y\right)=0$
$\Rightarrow\left(\lambda^2-2 \lambda+2\right) x^2+(4 x y-6 x y \lambda)$ $\left(\lambda^2-4 \lambda+2\right) y^2=0$
$\Rightarrow\left(\lambda^2-2 \lambda+2\right) x^2+(4-6 \lambda) x y$
$+\left(\lambda^2-4 \lambda+2\right) y^2=0$
Now, $\angle A O B=90^{\circ}$ where $O$ is the origin.
Condition for $\angle A D B=90^{\circ}$ is given by;
$\left(\right.$ Coefficient of $\left.x^2\right)+\left(\right.$ Coefficient of $\left.y^2\right)=0$
$\Rightarrow \quad\left(\lambda^2-2 \lambda+2\right)+\left(\lambda^2-4 \lambda+2\right)=0$
$2 \lambda^2-6 \lambda+4=0$
$\Rightarrow \quad \lambda^2-3 \lambda+2=0$
$\Rightarrow \quad(\lambda-1)(\lambda-2)=0$
$\Rightarrow \quad \lambda=1,2$
Hence, option (a) is correct.