I was reading N. Hitchin's lecture notes Chapter 4 The Klein programme. At the beginning of page 67, Hitchin wrote that
We consider the subset $D\subset P^2(\mathbf{R})$ defined by $$ D = \{[a,b,c]\in P^2(\mathbf{R}): b^2-4ac<0\}. $$ If $a+c=0$, then $b^2-4ac>0$ so $D$ lies in the complement of the line $a+c=0$ in $P^2(\mathbf{R})$ and we can regard $D$ as the interior of the circle $(b/(a+c))^2+((a-c)/(a+c))^2 = 1$ in $\mathbf{R}^2$. The group $PGL(2,\mathbf{R})$, and its subgroup $PSL(2,\mathbf{R})$, acts naturally on this space of quadrics.
Hitchin said that historically, $D$ is the Klein-Beltrami model.
Question: Could someone explain how the group $PSL(2,\mathbf{R})$ acts naturally on $D$? I am thinking that maybe Hitchin means that $PSL(2,\mathbf{R})$ preserves the interior of the unit circle in $\mathbf{R}^2$. But this is clearly not true.
As discussed at the bottom of page 66, a point $[a,b,c]\in D$ can be thought of as the equation for a quadric $$ax_0^2+bx_0x_1+cx_1^2$$ in $\mathbb{P}^1(\mathbb{R})$. The group $PGL(2,\mathbb{R})$ acts on the set of such quadrics by acting on the variables $x_0$ and $x_1$: an element of $PGL(2,\mathbb{R})$ gives a linear substitution to make for $x_0$ and $x_1$ which transforms the quadric into a different quadric. Or, in geometric terms, $PGL(2,\mathbb{R})$ acts on $\mathbb{P}^1(\mathbb{R})$ and maps the subscheme defined by one quadric to the subscheme defined by another quadric.