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Terris Becker edited untitled.tex
about 8 years ago
Commit id: fc732b54185081bcebf9737ad8d345957cb05a6f
deletions | additions
diff --git a/untitled.tex b/untitled.tex
index 22fe4dd..feaeb36 100644
--- a/untitled.tex
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&\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}+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}+b\\
&\iff(f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}(d^{2}+e^{2})+b\\ (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
$r=\sqrt{\frac{1}{4}(d^{2}+e^{2})+b}$, $r=\sqrt{\frac{1}{4}(d^{2}+e^{2})-b}$, 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}