<p>A rectangle of maximum area is inscribed in the circle \(|z - 3 - 4i| = 1\). If one vertex of the rectangle is \(4 + 4i\), then another adjacent vertex of this rectangle can be</p>
Step-by-Step Solution
Key Concept: For a rectangle inscribed in a circle with center C, the diagonals pass through C and are equal in length. If one vertex is given, use the center to find the diametrically opposite vertex, then exploit the perpendicularity condition for adjacent vertices on the circle.
<p><strong>Step 1:</strong> Circle center is C = 3 + 4i with radius r = 1. Given vertex A = 4 + 4i.</p><p><strong>Step 2:</strong> Verify A is on circle: |4 + 4i - 3 - 4i| = |1| = 1 ✓</p><p><strong>Step 3:</strong> For a rectangle inscribed in a circle, diagonals are diameters. If A = 4 + 4i is one vertex, the diagonally opposite vertex is: A' = 2C - A = 2(3 + 4i) - (4 + 4i) = 6 + 8i - 4 - 4i = 2 + 4i</p><p><strong>Step 4:</strong> For maximum area rectangle in a circle, it must be a square (when all inscribed rectangles have equal diagonal). Adjacent vertices B and D to A must satisfy: (B - C) ⊥ (A - C) and |B - C| = |A - C| = 1.</p><p><strong>Step 5:</strong> Vector from C to A: (4 + 4i) - (3 + 4i) = 1. A perpendicular vector of same length is ±i. So B = C ± i·(A - C) = (3 + 4i) ± i(1) = 3 + 4i ± i</p><p><strong>Step 6:</strong> Adjacent vertices: B = 3 + 5i or B = 3 + 3i</p><p>∴ Answer: B</p>
Correct Answer: B