The equation of the sphere passing through th | vedclass

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(A) Let the points be $A(1,0,0), B(0,1,0)$ and $C(1,1,1)$.Calculate the distances between the points:$AB = \sqrt{(0-1)^2 + (1-0)^2 + 0^2} = \sqrt{2}$$

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$3(x^2+y^2+z^2)-4x-4y-2z+1=0$

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(A) Let the points be $A(1,0,0), B(0,1,0)$ and $C(1,1,1)$.Calculate the distances between the points:$AB = \sqrt{(0-1)^2 + (1-0)^2 + 0^2} = \sqrt{2}$$BC = \sqrt{(1-0)^2 + (1-1)^2 + (1-0)^2} = \sqrt{2}$$CA = \sqrt{(1-1)^2 + (1-0)^2 + (1-0)^2} = \sqrt{2}$Since $AB = BC = CA = \sqrt{2}$,the points form an equilateral triangle.The sphere with the smallest radius passing through these points has its center at the centroid of the triangle $ABC$.Center $C' = \left(\frac{1+0+1}{3}, \frac{0+1+1}{3}, \frac{0+0+1}{3}\right) = \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)$.The radius $R$ is the distance from $C'$ to any point,say $A(1,0,0)$:$R^2 = \left(\frac{2}{3}-1\right)^2 + \left(\frac{2}{3}-0\right)^2 + \left(\frac{1}{3}-0\right)^2 = \frac{1}{9} + \frac{4}{9} + \frac{1}{9} = \frac{6}{9} = \frac{2}{3}$.The equation of the sphere is $(x-\frac{2}{3})^2 + (y-\frac{2}{3})^2 + (z-\frac{1}{3})^2 = \frac{2}{3}$.Expanding this: $x^2 - \frac{4x}{3} + \frac{4}{9} + y^2 - \frac{4y}{3} + \frac{4}{9} + z^2 - \frac{2z}{3} + \frac{1}{9} = \frac{6}{9}$.$x^2 + y^2 + z^2 - \frac{4x}{3} - \frac{4y}{3} - \frac{2z}{3} + \frac{9}{9} = \frac{6}{9}$.$x^2 + y^2 + z^2 - \frac{4x}{3} - \frac{4y}{3} - \frac{2z}{3} + 1 = \frac{2}{3}$.Multiplying by $3$: $3(x^2+y^2+z^2) - 4x - 4y - 2z + 3 = 2$.$3(x^2+y^2+z^2) - 4x - 4y - 2z + 1 = 0$.

The equation of the sphere passing through the points $(1,0,0), (0,1,0)$ and $(1,1,1)$ and having the smallest radius is:

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