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. $c^{2} - 4 = 0$
• B. $c^{2} - 8 = 0$
• C. $c^{2} - 9 = 0$
• D. $c^{2} - 10 = 0$

Answer 1

Answer: (C)

$c^{2} - 9 = 0$

Answer 2

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

⇒$x^{2} + y^{2} + ( - 2)\left( \frac{2\sqrt{2}x - y}{- c} \right)^{2} = 0$

⇒ $x^{2} + y^{2} - \frac{2}{c^{2}}\left( 8x^{2} + y^{2} - 4\sqrt{2}xy \right) = 0$

⇒ $x^{2}\left( 1 - \frac{16}{c^{2}} \right) + y^{2}\left( 1 - \frac{2}{c^{2}} \right) + \frac{8\sqrt{2}xy}{c^{2}} = 0$

If these lines are perpendicular, $1 - \frac{16}{c^{2}} + 1 - \frac{2}{c^{2}} = 0$

$\Rightarrow$ $\frac{2c^{2} - 18}{c^{2}} = 0$ ⇒ $c^{2} - 9 = 0$.