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(A) The equation of the pair of lines passing through the origin and the intersection of $x^2+y^2=4$ and $x+y=a$ is obtained by homogenizing the circl

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$2$

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(A) The equation of the pair of lines passing through the origin and the intersection of $x^2+y^2=4$ and $x+y=a$ is obtained by homogenizing the circle equation using the line equation: $x^2+y^2=4 \left(\frac{x+y}{a}\right)^2$ $a^2(x^2+y^2)=4(x^2+y^2+2xy)$ $(a^2-4)x^2 - 8xy + (a^2-4)y^2 = 0$ Since the lines are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(a^2-4) + (a^2-4) = 0$ $2(a^2-4) = 0$ $a^2 = 4$ Since $a>0$,we have $a=2$.

$A$ pair of perpendicular lines passes through the origin and also through the points of intersection of the curve $x^2+y^2=4$ with $x+y=a$,where $a>0$. Then $a$ is equal to

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