The angle between the pair of straight lines formed by joining the points of intersection of $x^2+y^2=4$ and…
The angle between the pair of straight lines formed by joining the points of intersection of $x^2+y^2=4$ and $y=3 x+c$ to the origin is a right angle. Then $c^2$ is equal to
$20$
$13$
$1 / 5$
$5$
Solution
Since, the angle is right angle.
$\therefore$ Homogenising, $x^2+y^2=4\left(\frac{y-3 x}{c}\right)^2$
$\Rightarrow \quad c^2\left(x^2+y^2\right)=4\left(y^2+9 x^2-6 x y\right)$
These lines are perpendicular.
$\therefore$ coefficient of $x^2+$ coefficient of $y^2=0$
$\begin{aligned} \Rightarrow & c^2-36+c^2-4 & =0 \\ \Rightarrow & 2 c^2 & =40 \\ \Rightarrow & c^2 & =20\end{aligned}$