Question

The pair of straight lines joining the origin to the points of intersection of the line $y=2\sqrt{2}x+c$ and the circle $x^2+y^2=2$ are at right angles, if

• A. $c^2-4$
• B. $c^2-8$
• C. $c^2-9$
• D. $c^2-10$

Answer 1

The correct option is C

$c^2-9$

Explanation for the correct option

Given equation of line is,

$$\begin{gathered} y=2\sqrt{2}x+c \\ \Rightarrow y-2\sqrt{2}x=c.......\left(i\right) \end{gathered}$$

Given equation of circle is,

$x^2+y^2=2.......\left(ii\right)$

We make $\left(i\right)$ a homogeneous equation by using $\left(i\right)$

$$\begin{gathered} \Rightarrow x^2+y^2-2{\left(1\right)}^2=0 \\ \Rightarrow x^2+y^2-2\frac{{\left(y-2\sqrt{2}x\right)}^2}{c^2}=0 \\ \Rightarrow c^2{x}^2+c^2{y}^2-2\left(y^2-4\sqrt{2}xy+8x^2\right)=0 \\ \Rightarrow x^2\left(c^2-16\right)+y^2\left(c^2-2\right)+8\sqrt{2}xy=0 \end{gathered}$$

The condition for the lines to be perpendicular is

$$\begin{gathered} \mathrm{coefficient}\mathrm{of}{x}^2+\mathrm{coefficient}\mathrm{of}{y}^2=0 \\ \Rightarrow c^2-16+c^2-2=0 \\ \Rightarrow 2c^2-18=0 \\ \Rightarrow c^2-9=0 \end{gathered}$$

Hence, option $\left(\mathrm{C}\right)$ is the correct answer.