<s>[INST] $ax^2 + 2hxy + by^2 +2gx + 2fy+c = 0$ represents a point when $\Delta = 0$ and $h^2 - ab < 0$. How can I show that $ax^2  + 2hxy + by^2 +2gx + 2fy+c = 0$  represents a point when  $\Delta = 0$  and $h^2 - ab < 0$.
I know I can do it by completing the square. But it turns out to be too tedious. Can anyone help me ?
 [/INST] We can write $$y=\frac{hx+f\pm \sqrt{(h^2-ab)x^2+2(hf-bg)x+(f^2-ac)}}{b}~~~~~(1)$$
For a pair of lines, $y$ needs to be real for all real values of $x$. inside square root we should have a perfect square of like $(px+q)^2>0$. For  a point it should be like $-(rx+s)^2$ ($y$ non real excepting one value of $x=-s/r$. For this we need to have  $A<0$ and $B^2-4AC=0$ for the quadratic in the radical. The two conditions give  $h^2-ab<0, (hf-bg)^2-(h^2-ab)(f^2-ac)=0=\Delta.~~~~(2)$
The examples are: $x^2+y^2=0, 3(x-1)^2+(y-x+2)^2=0, 3(x-1)^2+2(y-2)^2=0.$
Note that if $x,y$ are real then sum of perfect squares cannot be zero. The only possibility is that both squares are simultaneously zero. the points are $(0,0),(1,-1),(1,2)$ are the points represented by the three examples, respectively. Upon expansion, for them we get $a,b,c,f,g,h$ satisfying both parts of (2).
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