<p>The circle for which the line joining the points \((am^2, 2am)\) and \(\left(\dfrac{a}{m^2}, -\dfrac{2a}{m}\right)\) is a diameter, is</p>
Step-by-Step Solution
Key Concept: If two points form a diameter of a circle, the center is their midpoint and the radius is half the distance between them. For a circle with diameter endpoints, use the property that any point P on the circle satisfies: (P - endpoint₁)·(P - endpoint₂) = 0.
<p><strong>Step 1:</strong> Let A = (am², 2am) and B = (a/m², -2a/m) be the diameter endpoints.</p><p><strong>Step 2:</strong> For any point P(x, y) on the circle, use the property that the angle in a semicircle is 90°: (x - am²)(x - a/m²) + (y - 2am)(y + 2a/m) = 0</p><p><strong>Step 3:</strong> Expand the first term: x² - x(am² + a/m²) + a²·m²·(1/m²) = x² - ax(m² + 1/m²) + a²</p><p><strong>Step 4:</strong> Expand the second term: y² + y(2am - 2a/m) - 2am·(2a/m) = y² + 2ay(m - 1/m) - 4a²</p><p><strong>Step 5:</strong> Combine: x² + y² - ax(m² + 1/m²) + 2ay(m - 1/m) + a² - 4a² = 0</p><p><strong>Step 6:</strong> Simplify: x² + y² - ax(m² + 1/m²) + 2ay(m - 1/m) - 3a² = 0</p><p>∴ Answer: A</p>
Correct Answer: A