<p>The equation of a plane is \(x + 2y + 2z + 7 = 0\) and the equation of a sphere is \(x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0\). The radius of the circle of intersection of the plane and sphere is:</p>
Step-by-Step Solution
Key Concept: The radius of the circle of intersection equals √(R² - d²), where R is the sphere's radius and d is the perpendicular distance from the sphere's center to the plane.
Step 1: Convert sphere equation to standard form by completing the square: x^2 + 2x + y^2 - 2y + z^2 - 4z = 19 (x + 1)^2 - 1 + (y - 1)^2 - 1 + (z - 2)^2 - 4 = 19 (x + 1)^2 + (y - 1)^2 + (z - 2)^2 = 25 Center: C(-1, 1, 2), Radius: R = 5 Step 2: Find perpendicular distance from center C(-1, 1, 2) to plane x + 2y + 2z + 7 = 0: d = |(-1) + 2(1) + 2(2) + 7|/√(1^2 + 2^2 + 2^2) d = |-1 + 2 + 4 + 7|/√9 = |12|/3 = 4 Step 3: Apply circle radius formula: r = √(R^2 - d^2) = √(25 - 16) = √9 = 3 ∴ Answer: C
Correct Answer: C