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

Class 11 Mathematics Pair of straight lines Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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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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TS EAMCET 2008 All TS EAMCET

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

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Pair of straight lines

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Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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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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