(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 as:$x^2+y^2-4\l

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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 as:$x^2+y^2-4\left(\frac{x+y}{a}\right)^2=0$Multiplying by $a^2$,we get:$a^2(x^2+y^2)-4(x^2+y^2+2xy)=0$$(a^2-4)x^2-8xy+(a^2-4)y^2=0$Since the pair of lines is perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(a^2-4)+(a^2-4)=0$$2(a^2-4)=0$$a^2=4$$a=\pm 2$Hence,the required set of $a$ is $\{-2, 2\}$.

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

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$A$ 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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