The radius of the circle in which the sphere | vedclass

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(C) The equation of the sphere is $x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2w

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(C) The equation of the sphere is $x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get the center $C = (-u, -v, -w) = (-1, 1, 2)$ and the radius $R = \sqrt{u^2 + v^2 + w^2 - d} = \sqrt{(-1)^2 + 1^2 + 2^2 - (-19)} = \sqrt{1 + 1 + 4 + 19} = \sqrt{25} = 5$.The perpendicular distance $p$ from the center $C(-1, 1, 2)$ to the plane $x + 2y + 2z + 7 = 0$ is given by $p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|-1 + 2 + 4 + 7|}{\sqrt{1 + 4 + 4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.The radius $r$ of the circle formed by the intersection is given by $r = \sqrt{R^2 - p^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$.

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

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