A pair of perpendicular straight lines passes through the origin and also through the point of intersection…
A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^2+y^2=4$ with $x+y=a$. The set containing the value of ' $a$ ' is
$\{-2,2\}$
$\{-3,3\}$
$\{-4,4\}$
$\{-5,5\}$
Solution
To make the given curves $x^2+y^2=4$ and $x+y=a$ homogeneous.
$
\begin{array}{rlrl}
& \therefore & x^2+y^2-4\left(\frac{x+y}{a}\right)^2 & =0 \\
& \Rightarrow & a^2\left(x^2+y^2\right)-4\left(x^2+y^2+2 x y\right) & =0 \\
\Rightarrow & x^2\left(a^2-4\right)+y^2\left(a^2-4\right)-8 x y & =0
\end{array}
$
Since, this is a perpendicular pair of straight lines.
$
\begin{array}{ll}
\therefore & a^2-4+a^2-4=0 \\
\Rightarrow & a^2=4 \Rightarrow a= \pm 2
\end{array}
$
Hence, required set of $a$ is $\{-2,2\}$