JEE Main · 2005 · 08 May 2005 · Offline · Straight Lines

The two diameters cut the circle into four sectors; if one sector has thrice the area of an adjacent one, their central angles are in the ratio $3:1$, and since two adjacent sectors fill a straight angle, $\theta +3\theta =180^{\circ }$ gives $\theta =45^{\circ }$ for the angle between the lines. For the pair $a{x}^2+2(a+b)xy+b{y}^2=0$, the angle satisfies $\tan \theta =\left|\frac{2\sqrt{(a+b{)}^2-ab}}{a+b}\right|$. Setting $\tan 45^{\circ }=1$ gives $4[(a+b{)}^2-ab]=(a+b{)}^2$, i.e. $3(a+b{)}^2=4ab$; expanding, $3a^2+6ab+3b^2=4ab$, which simplifies to $3a^2+2ab+3b^2=0$.