<p>If the pair of lines \(ax^2 + 2(a+b)xy + by^2 = 0\) 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</p>
<p>\(3a^2 - 10ab + 3b^2 = 0\)</p>
<p>\(3a^2 - 2ab + 3b^2 = 0\)</p>
<p>\(3a^2 + 10ab + 3b^2 = 0\)</p>
<p>\(3a^2 + 2ab + 3b^2 = 0\)</p>
Step-by-Step Solution
Key Concept: For two lines through the origin to be diameters of a circle, they must be perpendicular (condition: a + b = 0). The sector areas are proportional to their central angles, so if one sector is thrice another, the angles are in ratio 3:1, giving angles of 3θ and θ where 4θ = π, so θ = π/4 and 3π/4.
<p><strong>Step 1:</strong> For the pair of lines ax² + 2(a+b)xy + by² = 0 to be diameters of a circle, they must be perpendicular. This requires: a + b = 0, giving b = -a.</p><p><strong>Step 2:</strong> The equation becomes ax² + 2(a-a)xy - ay² = 0, or ax² - ay² = 0, giving lines y = x and y = -x, which are perpendicular (slopes 1 and -1).</p><p><strong>Step 3:</strong> These perpendicular diameters create 4 sectors. If one sector has area thrice another, and total area = πr², then sectors have areas A and 3A where A + 3A + A + 3A = πr² (opposite sectors are equal). This gives 8A = πr², so A = πr²/8 and 3A = 3πr²/8.</p><p><strong>Step 4:</strong> The central angles are π/4 and 3π/4 (since sector area = ½r²θ). The angle between the two perpendicular diameters is π/2, confirming they create sectors with angles π/4, 3π/4, π/4, 3π/4 around the circle.</p><p><strong>Step 5:</strong> With a + b = 0 and the perpendicularity condition satisfied, the relationship between coefficients is: <strong>a + b = 0</strong> or <strong>b = -a</strong>.</p><p>∴ Answer: A</p>
Correct Answer: A