<p>If a circle C passing through the point (4, 0) touches the circle \(x^2 + y^2 + 4x - 6y = 12\) externally at the point (1, −1), then the radius of C is __________.</p>
Step-by-Step Solution
Key Concept: Use the collinearity property: the centers of two externally tangent circles and their point of tangency are collinear. The center of C lies on the line through (1, -1) and the center of the given circle, at a distance equal to C's radius from (1, -1).
<p><strong>Step 1:</strong> Rewrite the given circle in standard form: x² + y² + 4x - 6y = 12 → (x + 2)² + (y - 3)² = 25. Center O₁ = (-2, 3), radius r₁ = 5.</p><p><strong>Step 2:</strong> For external tangency at (1, -1), the center C of circle C lies on the line through O₁(-2, 3) and (1, -1). Direction vector: (1-(-2), -1-3) = (3, -4).</p><p><strong>Step 3:</strong> Parametric form of the line: (x, y) = (1, -1) + t(3, -4). Let C = (1 + 3t, -1 - 4t) with radius r_C. Distance from C to (1, -1) equals r_C: √[(3t)² + (-4t)²] = r_C → 5|t| = r_C.</p><p><strong>Step 4:</strong> For external tangency: |CO₁| = r_C + 5. Distance from (1 + 3t, -1 - 4t) to (-2, 3): √[(3t + 3)² + (-4t - 4)²] = r_C + 5 → √[25(t + 1)²] = r_C + 5 → 5|t + 1| = r_C + 5.</p><p><strong>Step 5:</strong> Taking t > 0 (moving away from O₁): 5t = r_C and 5(t + 1) = r_C + 5 both give consistent result. From first: r_C = 5t. From constraint that C passes through (4, 0): (4 - 1 - 3t)² + (0 + 1 + 4t)² = (5t)² → (3 - 3t)² + (1 + 4t)² = 25t² → 9 - 18t + 9t² + 1 + 8t + 16t² = 25t² → 10 - 10t = 0 → t = 1.</p><p><strong>Step 6:</strong> Therefore r_C = 5(1) = 5.</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: 5