Question

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

• A. 5
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (D)

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