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\begin{align}  &\iff (f^{2}+g^{2})+(fd-eg)+b=0\\  &\iff f^{2}+fd+g^{2}-eg+b=0\\  &\iff (f+\frac{1}{2}d)^{2}-\frac{1}{4}d^{2}+(g-\frac{1}{2})^{2}-\frac{1}{4}e^{2}=0 (f+\frac{1}{2}d)^{2}-\frac{1}{4}d^{2}+(g-\frac{1}{2})^{2}-\frac{1}{4}e^{2}+b=0  \text{ (By completing the square)}\\ &\iff (f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}d^{2}+\frac{1}{4}e^{2}\\  &\iff(f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}(d^{2}+e^{2})\\ (f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}d^{2}+\frac{1}{4}e^{2}+b\\  &\iff(f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}(d^{2}+e^{2})+b\\  \end{align}  The above equation is a circle with radius $r=\sqrt{\frac{1}{4}(d^{2}+e^{2})}$, $r=\sqrt{\frac{1}{4}(d^{2}+e^{2})+b}$,  so $r^{2}$ must be greater than or equal to zero in order for a solution to exist. Thus, \begin{align}  \frac{1}{4}(d^{2}+e^{2})\geq0  \end{align}