<p>Let L = 0 be a common normal to the circle \(x^2 + y^2 - 2\alpha x - 36 = 0\) and the curve \(S : (1+x)^y + e^{xy} = y\) drawn at a point \(x = 0\) on S, then the radius of the circle is</p>
Step-by-Step Solution
Key Concept: Find the slope of the tangent to curve S at x=0 using implicit differentiation, then use the fact that the common normal to both curves must be perpendicular to the tangent at that point. The normal line's slope combined with the circle's geometry determines α and hence the radius.
<p><strong>Step 1:</strong> Find the point on curve S at x=0. Substituting x=0 into (1+x)^y + e^(xy) = y: (1)^y + e^0 = y, so 1 + 1 = y, giving y = 2. Point is (0, 2).</p><p><strong>Step 2:</strong> Differentiate S implicitly: (1+x)^y[ln(1+x)·dy/dx + y/(1+x)] + e^(xy)[y + x·dy/dx] = dy/dx. At (0,2): 1·[0 + 2] + 1·[2 + 0] = dy/dx, so dy/dx = 4. Tangent slope = 4, normal slope = -1/4.</p><p><strong>Step 3:</strong> The normal line at (0,2) with slope -1/4 is: y - 2 = -1/4(x - 0), or x + 4y - 8 = 0. So L: x + 4y - 8 = 0.</p><p><strong>Step 4:</strong> For this to be a normal to circle x² + y² - 2αx - 36 = 0, the line must pass through the center (α, 0). Substituting: α + 4(0) - 8 = 0, giving α = 8.</p><p><strong>Step 5:</strong> Circle equation: x² + y² - 16x - 36 = 0. Complete the square: (x-8)² + y² = 64 + 36 = 100. Radius = √100 = <strong>10</strong>.</p><p>∴ Answer: A</p>
Correct Answer: A