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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Solution

The correct option is C $3$
Given equation of sphere is $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ .
Centre of the sphere is $(- 1, 1, 2)$ and its radius is $\sqrt{1 + 1 + 4 + 19} = 5$
Length of the perpendicular from the centre on the plane is $∣ ∣ ∣ \frac{- 1 + 2 + 4 + 7}{\sqrt{1 + 4 + 4}} ∣ ∣ ∣ = 4$
Radius of the required circle is $\sqrt{5^{2} - 4^{2}} = 3$ .

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Similar questions

x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0

x + 2 y + 2 z + 7 = 0

x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0

x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0 = x + 2 y + 2 z + 7

x + 2 y - z = 4

x^{2} + y^{2} + z^{2} - x + z - 2 = 0

x^{2} + y^{2} = 4; z = 0

x + 2 y + 2 z = 0

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