The radius of the circle given by $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0 = x + 2 y + 2 z + 7$ , is

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Solution

The correct option is B 3
Radius of the sphere is $r = \sqrt{1^{2} + (- 1)^{2} + (- 2)^{2} + 19} = 5$
And center of the sphere is $C = (- 1, 1, 2)$
Now distance of the given plane from the centre is $C P = \frac{- 1 + 2 + 4 + 7}{\sqrt{1 + 4 + 4}} = 4$
Hence radius of the circle is $= \sqrt{r^{2} - C P^{2}} = 3$

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Similar questions

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 = 0

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

x + 2 y + 2 z = 15

x^{2} + y^{2} + z^{2} - 2 y - 4 z = 11

x^{2} + y^{2} + z^{2} - 3 x + 4 y - 2 z - 5 = 0

5 x - 2 y + 4 z + 7 = 0

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