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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 $u=1, v=-

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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 $u=1, v=-1, w=-2, d=-19$. The centre of the sphere is $(-u, -v, -w) = (-1, 1, 2)$. The radius of the sphere $R$ is given by $\sqrt{u^2+v^2+w^2-d} = \sqrt{1^2+(-1)^2+(-2)^2-(-19)} = \sqrt{1+1+4+19} = \sqrt{25} = 5$. Now,the perpendicular distance $p$ from the centre $(-1, 1, 2)$ 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 circumference of 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 formed 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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