$2x^2 + 2y^2 + 2z^2 = 8x - 24z + 1$
So here's my guess:
$2x^2 - 8x + 2y^2 + 2z^2 + 24z = 1$ Get all variables on one side
$2x^2 - 8x + 16 + 2y^2 + 2z^2 + 24z + 144 = 1 + 16 + 144$ Complete square
$(2x - 8)^2 + 2y^2 + (2z^2 + 24)^2 = 161$ Stopped here
Is it possible to complete the square with the $y$ value too? Would it just be 1?
$\endgroup$ 12 Answers
$\begingroup$Here's a way to do it: divide through by $2$ to get$$ x^2+y^2+z^2=4x-12z+\frac{1}{2}$$$$ x^2-4x+y^2+z^2 +12 z=\frac{1}{2}.$$Now, to complete the square for $x^2-4x$ we need to add $4$ to both sides. To complete the square for $z^2+12z$ we need to add $36$ to both sides. In total we add $40$ to both sides to get$$ x^2-4x+4+y^2+z^2+12z+36=\frac{81}{2}$$$$ (x-2)^2+y^2+(z+6)^2=\frac{81}{2}.$$As we know, this is the equation for a sphere of radius $9/\sqrt{2}$ centred at $(2,0,-6)$.
$\endgroup$ $\begingroup$Note that $(2x)^2 \ne 2x^2$. The answer is:$$2(x-2)^2 + 2(y-0)^2 + 2(z+6)^2 = 8 + 0 + 72 + 1$$Thus, center $(2,0,-6)$, and radius $\sqrt{81/2}$.
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