Question

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

• A. 5
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (D)

Homogenise the pair of lines
$x^2+y^2-2x-4y+2=0$ through $x+y=k$
we get
$x^2+y^2-2x\left(\frac{x+y}{k}\right)-4y\left(\frac{x+y}{k}\right)$
$+2{\left(\frac{x+y}{k}\right)}^2=0$
Since, intersection points of line and pair of lines make an angle $90^{\circ }$ at origin $O$.
$\therefore$ Coefficient of $x^2+$ Coefficient of $y^2=0$
$\Rightarrow \left(1-\frac{2}{k}+\frac{2}{k^2}\right)+\left(1-\frac{4}{k}+\frac{2}{k^2}\right)=0$
$\Rightarrow 2-\frac{6}{k}+\frac{4}{k^2}=0$
$\Rightarrow k^2-3k+2=0$
$\Rightarrow (k-2)(k-1)=0$
$\Rightarrow k=1,2$
But $k>1$
$\therefore k=2$