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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 by thanvigandhi_1

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