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(A) To make the given curves $x^2+y^2=4$ and $x+y=a$ homogeneous,we write the equation of the pair of lines passing through the origin and the interse

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$\{-2, 2\}$

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(A) To make the given curves $x^2+y^2=4$ and $x+y=a$ homogeneous,we write the equation of the pair of lines passing through the origin and the intersection points as follows:$x^2+y^2-4\left(\frac{x+y}{a}\right)^2=0$$a^2(x^2+y^2)-4(x^2+y^2+2xy)=0$$x^2(a^2-4) - 8xy + y^2(a^2-4)=0$Since this represents a pair of perpendicular straight lines,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(a^2-4) + (a^2-4) = 0$$2(a^2-4) = 0$$a^2 = 4$$a = \pm 2$Hence,the required set of $a$ is $\{-2, 2\}$.

$A$ 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

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