<p>The radius of the circle passing through the point (6, 2) and having \(x + y = 6\) as a normal and \(x + 2y = 4\) as a diameter is</p>
Step-by-Step Solution
Key Concept: A normal to a circle is perpendicular to the tangent and passes through the center. Use the fact that the center lies on the normal line, and the diameter endpoints satisfy the given diameter equation to find the center, then calculate radius using the given point.
<p><strong>Step 1:</strong> The normal to the circle is x + y = 6. The normal passes through the center, so the center C lies on x + y = 6. Let C = (h, k) where h + k = 6.</p><p><strong>Step 2:</strong> The diameter lies on x + 2y = 4. The center is the midpoint of any diameter, so the center must satisfy this diameter equation: h + 2k = 4.</p><p><strong>Step 3:</strong> Solve the system:<br>h + k = 6<br>h + 2k = 4<br>Subtracting: -k = 2, so k = -2 and h = 8.<br>Center C = (8, -2).</p><p><strong>Step 4:</strong> The circle passes through point P(6, 2). The radius is the distance from C to P:<br>r = √[(8-6)² + (-2-2)²] = √[4 + 16] = √20 = 2√5</p><p>∴ Answer: B</p>
Correct Answer: B