The plane x + 2y - z = 4 cuts the sphere x^2 | vedclass

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(D) The given equation of the sphere is $x^2 + y^2 + z^2 - x + z - 2 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz

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(D) The given equation of the sphere is $x^2 + y^2 + z^2 - x + z - 2 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get $u = -1/2$,$v = 0$,$w = 1/2$,and $d = -2$.The center of the sphere is $(-u, -v, -w) = (1/2, 0, -1/2)$.The radius of the sphere $R$ is given by $\sqrt{u^2 + v^2 + w^2 - d} = \sqrt{(-1/2)^2 + 0^2 + (1/2)^2 - (-2)} = \sqrt{1/4 + 0 + 1/4 + 2} = \sqrt{1/2 + 2} = \sqrt{5/2}$.The perpendicular distance $P$ from the center $(1/2, 0, -1/2)$ to the plane $x + 2y - z - 4 = 0$ is:$P = \frac{|(1/2) + 2(0) - (-1/2) - 4|}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{|1/2 + 1/2 - 4|}{\sqrt{6}} = \frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{9}{6}} = \sqrt{\frac{3}{2}}$.The radius $r$ of the circle formed by the intersection is given by $r = \sqrt{R^2 - P^2}$.$r = \sqrt{\frac{5}{2} - \frac{3}{2}} = \sqrt{\frac{2}{2}} = \sqrt{1} = 1$.

The plane $x + 2y - z = 4$ cuts the sphere $x^2 + y^2 + z^2 - x + z - 2 = 0$ in a circle of radius:

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