Question

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

• A. $\\{-2, 2\\}$
• B. $\\{-3, 3\\}$
• C. $\\{-4, 4\\}$
• D. $\\{-5, 5\\}$

Answer 1

Answer: (A)

$\\{-2, 2\\}$

Answer 2

To make the given curves $x^{2}+y^{2}=4$
and $x+y=a$ homogenous.
$\therefore x^{2}+y^{2}-4 \frac{x +y^{2}}{x}=0$
$\Rightarrow a^{2}\left(x^{2}+y^{2}\right)-4\left(x^{2}+y^{2}+2 x y\right)=0$
$\Rightarrow x^{2}\left(a^{2}-4\right)+y^{2}\left(a^{2}-4\right)-8 x y=0$
Since, this is a perpendicular pair of straight lines.
$\therefore a^{2}-4+a^{2}-4=0$
$\Rightarrow a^{2}=4$
$\Rightarrow a=\pm 2$
Hence, required set of a is $\\{-2,2\\}$.