# Solving a quadratic system of equations vs solving them linearly

How can I solve a system of these equations in $A$, $B$, $C$ and $D$?

$$A^2(y^2+z^2)+B^2(x^2+z^2)+C^{\,2}(x^2+y^2)-2ABxy-2BCyz-2CAzx - D=0$$

Here, $x$, $y$, $z$ represent known points, and they are not unknowns. Should I take 7 points to generate a system of 7 linear equations to solve for $A^2$, $B^2$, $C^{\,2}$, $AB$, $BC$, $CA$ and $D$? Or, I only need 4 points in this quadratic system in $A$, $B$ and $C$?

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What do you mean by points? Are they real numbers, or coordinates in 2-D / 3-D? – Calvin Lin Feb 20 '13 at 11:41
They are coordinates in 3-d, but they are known coordinates. A,B,C and D are unknowns. – Sonu Jha Feb 20 '13 at 11:45