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If $A (-1 , 4 , -3 )$ is one end of a diameter $AB$ of the sphere $x^{2}+y^{2}+z^{2}-3x-2y+2z-15=0$ than, Find the coordinates of $B$.

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• General equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ centre $(-u, -v, -w)$ radius $= \sqrt{u^2+v^2+w^2-d}$
The centre of the sphere $x^2+y^2+z^2-3x-2y+2z-15=0$ is at $(-u, -v, -w)$ where $2u=-3, \: 2v=-2, \: 2w=2.$
$\therefore$ the centre is at $C \bigg( \large\frac{3}{2}, 1, -1 \bigg)$
$C$ is the midpoint of the diameter $AB$, where $A(-1, 4, -3)\: and \: B(x_1, y_1, z_1). \therefore \: \large\frac{-1+x_1}{2}=\large\frac{3}{2}, \large\frac{4+y_1}{2}=1\: \large\frac{-3+z_1}{2}=-1$ i.e., $x_1=4,\: y_1=-2, \: z_1=1$
The centre is at $(4, -2, 1)$
answered Jun 18, 2013

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