Q27 / 50 · Manipal (MET) 2024 · Mathematics
If 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
- Correct answerA$20$
- B$13$
- C$\frac{1}{5}$
- D$5$
Solution
$\begin{aligned}
& \because x^2+y^2=4 \Rightarrow x^2+y^2-4(1)^2=0 ...(i)\\
& \text { and } y=3 x+c \Rightarrow \frac{y-3 x}{c}=1
\end{aligned}$
Put the value of ' 1 ' in Eq. (i)
$\begin{aligned}
& \Rightarrow x^2+y^2-4\left(\frac{y-3 x}{c}\right)^2=0 \\
& \Rightarrow x^2+y^2-4\left(\left(\frac{y}{c}\right)^2+\left(\frac{3 x}{c}\right)^2-\frac{6 x y}{c^2}\right)=0 \\
& \Rightarrow x^2\left(1-\frac{36}{c^2}\right)+y^2\left(1-\frac{4}{c^2}\right)+\frac{24 x y}{c^2}=0
\end{aligned}$
According to the question,
$\begin{aligned}
& \left(1-\frac{36}{c^2}\right)+\left(1-\frac{4}{c^2}\right)=0 \Rightarrow 2=\frac{40}{c^2} \\
\Rightarrow & \quad c^2=20
\end{aligned}$
Hence, the correct answer is option: 'A'.