TITLE: Show that there is no affine transformation that takes a circle to a hyperbola in $\mathbb{R}^2$
QUESTION [7 upvotes]: "Show that the standard circle (defined by $f(x,y) = x^2 + y^2 - 1$) is not equivalent to the standard hyperbola (defined by $g(x,y) = x^2 - y^2 - 1$). That is, show that there is no $[A,\overline{s}] \in \text{Aff}(\mathbb{R}^2)$ such that $[A,\overline{s}] \cdot f(x,y) = g(x,y)$. Check that there is such an $[A,\overline{s}]$ if we allow $A \in \text{GL}_2(\mathbb{C}).$"
I've reduced this to showing that there are no $a,b,c,d,s,t \in \mathbb{R}$ such that $$f(ax+as+by+bt,\: cx + cs + dy+dt) = g(x,y).$$
$$\Rightarrow(ax+as+by+bt)^2+(cx + cs + dy+dt)^2 - 1=x^2-y^2-1$$
How should I proceed? Expanding that expression probably isn't the best way to do it.

REPLY [5 votes]: Why think when you can simply compute?
So $f(x,y)=x^2+y^2-1$. Consider the affine transformation given by $$\left\{\begin{array}{l}x\leftarrow ax+by+s\\y\leftarrow cx+dy+t\end{array}\right.$$ Applying it to $f$ we get $$f(ax+by+s, cx+dy+t)=(a^2+c^2) x^2+2(a b + c d)xy+(b^2 + d^2)y^2+\text{terms of lower degree}.$$ If this is to be equal to $g(x,y)=x^2-y^2-1$, then, in particular, looking at the coefficient of $y^2$ we see that we must have $$b^2+d^2=-1.$$ Of course, this is not going to work...

REPLY [2 votes]: Let us call $f_0$, $f_1$ and $f_2$ the parts of $f$ which are homogeneous of degree $0$, $1$ and $2$, respectively, so that in particular $f=f_2+f_1+f_0$, and similarly for $g$.
If $[A,s]$ is an invertible affine transformation and $[A,s]\cdot f=g$, then one checks easily that $[A,s]\cdot f_i=g_i$ for each $i\in\{0,1,2\}$. In particular, $$[A,s]\cdot(x^2+y^2)=x^2-y^2.$$
But two polynomials which are affinely equivalent are either both irreducible or both reducible, yet $x^2+y^2$ is irreducible and $x^2-y^2$ is not.
This argument is «geometric»: we are using the fact that a circle has no asymptotic directions while the hyperbola has two, and that the property of having asymtotic directions is preserved under affine equivalence.