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: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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The radius of the circle in which the sphere x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0 is cut by the plane x + 2 y + 2 z + 7 = 0 is:

Solution & Step-by-step Explanation

Sphere: x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0 . Center C = (- 1, 1, 2) . Radius R = (- 1)^{2} + 1^{2} + 2^{2} - (- 19) = 1 + 1 + 4 + 19 = 25 = 5 . Perpendicular distance from C to the plane x + 2 y + 2 z + 7 = 0 : d = \frac{∣ - 1 + 2 ( 1 ) + 2 ( 2 ) + 7∣}{1 ^{2} + 2 ^{2} + 2 ^{2}} = \frac{∣ - 1 + 2 + 4 + 7∣}{3} = \frac{12}{3} = 4 Radius of the circle r = R^{2} - d^{2} = 5^{2} - 4^{2} = 25 - 16 = 9 = 3 .

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The radius of the circle in which the sphere x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0 is cut by the plane x + 2 y + 2 z + 7 = 0 is:

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