<s>[INST] Why does the equation of a circle have to have the same $x^2,y^2$ coefficients? In one of my geometry texts, it tells me they should be the same but not why. I am unsatisfied with this.
Suppose that:
$$ax^2+by^2 + cx + dy + f = 0 \text{ such that } a \neq b$$
is the equation of some circle. Upon completing the square and rearranging, I obtain
$$a\left(x+\frac{c}{2a}\right)^2 + b\left(y+\frac{d}{2b}\right)^2 = \frac{c^2}{4a} + \frac{d^2}{4b} - f$$
I know that a circle is defined as the set of points a fixed distance from a fixed point.
How can i arrive at a satisfying contradiction? At the moment I just can't see it. 
 [/INST] If $a\ne b$ and both of them are non-zero, then your equation above
$$\begin{align*}
a\left(x+\frac{c}{2a}\right)^2 + b\left(y+\frac{d}{2b}\right)^2 &= \frac{c^2}{4a} + \frac{d^2}{4b} - f\\
\left(x+\frac{c}{2a}\right)^2 + \frac ba\left(y+\frac{d}{2b}\right)^2 &= \frac{c^2}{4a^2} + \frac{d^2}{4ab} - \frac fa\\
\left(x+\frac{c}{2a}\right)^2 + \left(y\sqrt{\frac ba}+\frac{d}{2b}\sqrt\frac ba\right)^2 &= \frac{c^2}{4a^2} + \frac{d^2}{4ab} - \frac fa\\
\end{align*}$$
If $\frac ba > 0$, this means the equation only represents a circle in the scaled $\left(x,y\sqrt{\frac ba}\right)$ plane. In other words, the equation represents an ellipse.
If $\frac ba < 0$, this is not a circle either.
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