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If the pair of lines $a x^2+2(a+b) x y+b y^2=0$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then
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The correct answer is:
$3 a^2+2 a b+3 b^2=0$
$3 a^2+2 a b+3 b^2=0$
$$
\begin{aligned}
& \left|\frac{2 \sqrt{(a+b)^2-a b}}{a+b}\right|=1 \\
& \Rightarrow(a+b)^2=4\left(a^2+b^2+a b\right) \\
& \Rightarrow 3 a^2+3 b^2+2 a b=0 .
\end{aligned}
$$
\begin{aligned}
& \left|\frac{2 \sqrt{(a+b)^2-a b}}{a+b}\right|=1 \\
& \Rightarrow(a+b)^2=4\left(a^2+b^2+a b\right) \\
& \Rightarrow 3 a^2+3 b^2+2 a b=0 .
\end{aligned}
$$
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