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The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 z - 2 y - 4 z - 19 = 0$ is cut by the plane x+2y+22+7=0 is

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Correct option is B

Center of the sphere is (-1, 1,2) and its radius = 1+1+4+19=5C L, perpendicular distance of C from plane, is −1+2+4+71+4+4=4Now AL2 = CA2 - CL2 =25 - 16=9Hence, radius of the circle 9=3

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The radius of the circle in which the sphere x2+y2+z2+2z−2y−4z−19=0 is cut by the plane x+2y+22+7=0 is

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The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 z - 2 y - 4 z - 19 = 0$ is cut by the plane x+2y+22+7=0 is