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(B) 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 centre

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$3$

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(B) 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 centre $C = (-u, -v, -w) = (-1, 1, 2)$.The radius of the sphere $R$ is given by $R = \sqrt{u^2+v^2+w^2-d} = \sqrt{(-1)^2 + 1^2 + 2^2 - (-19)} = \sqrt{1+1+4+19} = \sqrt{25} = 5$.The perpendicular distance $p$ from the centre $C(-1, 1, 2)$ to the plane $x+2y+2z+7=0$ is:$p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|-1 + 2 + 4 + 7|}{\sqrt{1+4+4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.Let $r$ be the radius of the circle. In the right-angled triangle formed by the centre of the sphere,the centre of the circle,and a point on the circle,we have $R^2 = p^2 + r^2$.$r^2 = R^2 - p^2 = 5^2 - 4^2 = 25 - 16 = 9$.Therefore,$r = \sqrt{9} = 3$.

The radius of the circle given by the intersection of the sphere $x^2+y^2+z^2+2x-2y-4z-19=0$ and the plane $x+2y+2z+7=0$ is:

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