A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^2+y^2=4withx+y=a$. The set containing the value of ‘a’ is

Given Equation

 $\begin{array}{l}{x}^2+y^2-4=0--\left(1\right)\\ andx+y=a\\ \Rightarrow \frac{x+y}{a}=1--\left(2\right)\end{array}$         

Homogenization of (1) w.r.to (2)

$\begin{array}{l}\Rightarrow x^2+y^2-4{\left(1\right)}^2=0\\ \Rightarrow x^2+y^2-4{\left(\frac{x+y}{a}\right)}^2=0\\ \Rightarrow x^2+y^2-4\left(\frac{x^2+y^2+2xy}{a^2}\right)=0\\ \Rightarrow x^2+y^2-\frac{4}{a^2}{x}^2-\frac{4}{a^2}{y}^2-\frac{8}{a^2}xy=0\end{array}$

$\begin{array}{l}Thelinesare\perp riff{x}^2\mathrm{co}eff+y^2\mathrm{co}eff=0\\ \Rightarrow 1+1-\frac{4}{a^2}-\frac{4}{a^2}=0\\ \Rightarrow 2-\frac{8}{a^2}=0\\ \Rightarrow 1-\frac{4}{a^2}=0\\ \Rightarrow a^2=4\Rightarrow a=\pm 2\end{array}$