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 ${}^{\prime }{a}^{\prime }$ is

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

Answer 1

The correct option is A $\{-2,2\}$

To make the given curves $x^2+y^2=4$ and $x+y=a$ homogenous.
Therefore, $x^2+y^2-4{\left(\frac{x+y}{a}\right)}^2=0$
$\Rightarrow a^2(x^2+y^2)-4(x^2+y^2+2xy)=0$
$\Rightarrow x^2(a^2-4)+y^2(a^2-4)-8xy=0$
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 $\{-2,2\}$.