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Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Mathematics3D-Coordinate GeometryPlaneMedium2 minQB

MathematicsMediumsingle choice

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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Answer:

Solution:

Centre of the sphere is (-1, 1, 2) and its radius is $1 + 1 + 4 + 19 = 5$ . Length of the perpendicular from the centre on the plane is $\frac{- 1 + 2 + 4 + 7}{1 + 4 + 4} = 4$ . Radius of the required circle is $5^{2} - 4^{2} = 3$ .

Stream:JEESubject:MathematicsTopic:3D-Coordinate GeometrySubtopic:Plane

⏱ 2mℹ️ Source: QB

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