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(A) Let the general equation of the sphere be $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$. Since the sphere passes through the origin $(0, 0, 0)$,we s

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$\left( \frac{1}{2}, 1, 2 \right)$

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(A) Let the general equation of the sphere be $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$. Since the sphere passes through the origin $(0, 0, 0)$,we substitute $x=0, y=0, z=0$ to get $d = 0$. Since it passes through $(0, 2, 0)$,we have $0^2 + 2^2 + 0^2 + 2u(0) + 2v(2) + 2w(0) + 0 = 0$,which simplifies to $4 + 4v = 0$,so $v = -1$. Since it passes through $(1, 0, 0)$,we have $1^2 + 0^2 + 0^2 + 2u(1) + 2v(0) + 2w(0) + 0 = 0$,which simplifies to $1 + 2u = 0$,so $u = -1/2$. Since it passes through $(0, 0, 4)$,we have $0^2 + 0^2 + 4^2 + 2u(0) + 2v(0) + 2w(4) + 0 = 0$,which simplifies to $16 + 8w = 0$,so $w = -2$. The centre of the sphere is given by $(-u, -v, -w)$. Substituting the values,the centre is $\left( -(-1/2), -(-1), -(-2) \right) = \left( \frac{1}{2}, 1, 2 \right)$.

The centre of the sphere passing through the four points $(0, 0, 0), (0, 2, 0), (1, 0, 0)$ and $(0, 0, 4)$ is

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