x^2+y^2+z^2=R^2

(x-a)^2+(y-b)^2+(z-c)^2=R^2

(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})+(z-z_{1})(z-z_{2})=0 \end{align} \begin{align} \text{where $P_{1}(x_{1},y_{1},z_{1})$, $P_{2}(x_{2},y_{2},z_{2})$ are the ends of a diameter.}

\begin{vmatrix} x^2+y^2+z^2 & x & y & z & 1 \\ x_{1}^2+y_{1}^2+z_{1}^2 & x_{1} & y_{1} & z_{1} & 1 \\ x_{2}^2+y_{2}^2+z_{2}^2 & x_{2} & y_{2} & z_{2} & 1 \\ x_{3}^2+y_{3}^2+z_{3}^2 & x_{3} & y_{3} & z_{3} & 1 \\ x_{4}^2+y_{4}^2+z_{4}^2 & x_{4} & y_{4} & z_{4} & 1 \\ \end{vmatrix} =0

Ax^2+Ay^2+Az^2+Dx+Ey+Fz+M=0 \qquad (A is nonzero) \end{align} \begin{align} \text{O centro da esfera tem coordenadas $(a,b,c)$, com} \end{align} \begin{align} a=-\frac{D}{2A} \qquad b=-\frac{E}{2A} \qquad c=-\frac{F}{2A} \end{align} \begin{align} \text{O raio da esfera é $R=\frac{\sqrt[]{D^2+E^2+F^2-4A^2M}}{2A}$}