Show that four non-coplanar points in $\mathbb{R}^3$ determine an unique sphere.
I have no idea how to solve this exercise. Thank you for your help.
$\endgroup$ 12 Answers
$\begingroup$Hint:
The equation of a sphere of center $C=(\alpha,\beta,\gamma)$ and radius $r$ is: $$ (x-\alpha)^2+(y-\beta)^2+(z-\gamma)^2=r^2 $$ that becomes $$ x^2+y^2+z^2+ax+by+cz+d=0 $$ with $$ a=-2\alpha \quad b=-2\beta \quad c=-2\gamma \quad d=\alpha^2+\beta^2+\gamma^2-r^2 $$
so if we have four points, substituting the coordinates of these points in the equation we find a linear system of four equations in four unknowns $a,b,c,d$ that has one solution if the four points are not coplanar (look at its determinant).
$\endgroup$ 5 $\begingroup$Hint.
Consider the perpendicular bisector planes of three segments joining couples of points. Those planes intersect at a point that is the center of a sphere passing through the four points. And this is the only sphere having such property.
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