<p>The radius of the circle in which the sphere \(x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0\) is cut by the plane \(x + 2y + 2z + 7 = 0\) is</p>
Step-by-Step Solution
Key Concept: The intersection of a sphere and plane forms a circle whose radius is found using r = √(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: Find the center and radius of the sphere Rewrite x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0 by completing the square: (x^2 + 2x + 1) + (y^2 - 2y + 1) + (z^2 - 4z + 4) - 1 - 1 - 4 - 19 = 0 (x + 1)^2 + (y - 1)^2 + (z - 2)^2 = 25 Center: C = (-1, 1, 2), Radius: R = 5 Step 2: Find the perpendicular distance from center to plane Plane equation: x + 2y + 2z + 7 = 0 Distance 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 the circle radius formula r = √(R^2 - d^2) = √(25 - 16) = √9 = 3 ∴ Answer: C
Correct Answer: C