<p>A circle of unit radius touches positive \(x\)-axis and positive \(y\)-axis at \(A\) and \(B\) respectively. A variable line passing through origin intersects the circle in two points \(D\) and \(E\). If the slope of the line is \(m\) and the area of triangle \(DEB\) is maximum then \(\dfrac{1}{m^2}\) is equal to:</p>
Step-by-Step Solution
Key Concept: The circle has center at (1,1) since it touches both positive axes with unit radius. Maximize the area of triangle DEB by recognizing that area = ½ × base × height, where the base DE is a chord through origin and height is the perpendicular distance from B(0,1) to the line. The distance from B to line y=mx is maximized when m relates optimally to the circle's geometry.
<p><strong>Step 1:</strong> Since the circle of unit radius touches positive x-axis at A and positive y-axis at B, the center is C(1,1). Thus A(1,0) and B(0,1).</p><p><strong>Step 2:</strong> The line through origin has equation y = mx. Points D and E lie on both this line and the circle (x-1)² + (y-1)² = 1.</p><p><strong>Step 3:</strong> Substituting y = mx into circle equation: (x-1)² + (mx-1)² = 1, which gives (1+m²)x² - (2+2m)x + 1 = 0.</p><p><strong>Step 4:</strong> The chord length DE can be found using the formula: |DE| = 2√(r² - d²), where d is distance from center C(1,1) to line mx - y = 0. Thus d = |m - 1|/√(1+m²).</p><p><strong>Step 5:</strong> Area of triangle DEB = ½ × |DE| × (perpendicular distance from B to line DE). The distance from B(0,1) to line y = mx is |1|/√(1+m²) = 1/√(1+m²).</p><p><strong>Step 6:</strong> Area = ½ × 2√(1 - (m-1)²/(1+m²)) × 1/√(1+m²) = √[(1+m²-m²+2m-1)/(1+m²)] × 1/√(1+m²) = √[2m/(1+m²)²].</p><p><strong>Step 7:</strong> To maximize, let f(m) = 2m/(1+m²)². Taking df/dm = 0: 2(1+m²)² - 2m·2(1+m²)·2m = 0, which simplifies to (1+m²) - 4m² = 0, giving 1 - 3m² = 0, so m² = 1/3.</p><p><strong>Step 8:</strong> Therefore, 1/m² = 3.</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3