Point circle from pair of straight line equation

In general, $$x^2+y^2+2gx+2fy+c=0$$ represents a circle with centre at $C(-g,-f)$. Equations of the form $$ax^2+2hxy+by^2=0$$ represents a pair of straight lines passing through origin. But the equation $2x^2+2y^2+xy=0$ represents just a point $(0,0)$.

When I came to know that it was not a pair of straight line equation I tried this: $$2x^2+2y^2+xy=2(x+y)^2-3xy=0$$ Using AM GM inequality, $(x+y)^2\ge 4xy$ $$2(x+y)^2\ge 8xy$$ $$3xy\ge 8xy$$ Which is only possible if $x=y=0$.

But is there an easy way to quickly check if the equation $ax^2+2hxy+by^2=0$ represents a point circle? After all, I used the above method only after knowing it's not a pair of straight lines.

• I think you might find an answer trying to find out the connection of the radius and $c$ having a closer look at your first equation. (Have not tried it, it's just the thought coming to my mind first.) compare it with $\left(x+g\right)^2+\left(y+f\right)^2=r$. – Max Oct 22 '15 at 13:14
• you can only use AM-GM if $x,y\geq 0$ – Dr. Sonnhard Graubner Oct 22 '15 at 13:15
• I think $2x^2+2y^2+xy=0$ is not a point circle , since for a circle $xy$ term should be zero. That is general equation of point circle is $(x+g)^2+(y+f)^2=0$. So the equation $2x^2+2^2+xy=0$ are Pair of Imaginary straight lines with Real point of Intersection $(0,0)$ – Ekaveera Kumar Sharma Oct 22 '15 at 13:31

Take $ax^2+2hxy+by^2=0$. Multiply by $4a$ $(a\neq 0)$ and complete the square to obtain the equivalent equation $$(2ax+hy)^2+(4ab-h^2)y^2=0$$

If $4ab-h^2\gt 0$ then both terms on the left-hand side are non-negative, and must therefore be zero. If $4ab-h^2=0$ you get the single line $2ax+hy=0$ (the two lines coincide to give a degenerate case). And if $4ab-h^2\lt 0$ you get two lines by factoring the left-hand side as the difference of two squares.

• It's perhaps worth saying that one can handle the case $a = 0$ by noting that the equation is unchanged under the replacement $(a, x) \leftrightarrow (b, y)$, so we can apply the same reasoning when $b \neq 0$. If $a = b = 0$, then the equation reduces to $h x y = 0$. If $h = 0$ the solution set consists of the entire plane, and if $h \neq 0$, the solution set is the union of the $x$- and $y$-axes. In each of these cases (except $a = b = h = 0$) the conclusion that the sign of $4 a b - h^2$ determines the character of the solution set still holds. – Travis Oct 22 '15 at 14:01
• @Travis I was just thinking whether I should add that - so thank you! – Mark Bennet Oct 22 '15 at 14:02
• You're welcome. And, anyway, the answer is quite nice (+1). – Travis Oct 22 '15 at 14:02

This is a special case of what is known as quadratic forms.

If you rewrite $ax^2+2hxy+by^2=0$ using matrices as $$\left[\matrix{x \\ y}\right]^T \left[\matrix{a & h \\ h & b}\right] \left[\matrix{x \\ y}\right] = 0$$ you can perform a change of variables to the inner matrix's eigenvectors to get something on the form $$\left[\matrix{x' \\ y'}\right]^T \left[\matrix{d_1 & 0 \\ 0 & d_2}\right] \left[\matrix{x' \\ y'}\right] = 0$$ which expands to $d_1{x'}^2 + d_2{y'}^2 = 0$

From this it is easy to see if it represents a single point, two intersecting lines or a single line just from the signs of $d_1$ and $d_2$ (or if one is zero).

If you instead had something non-zero on the right, you could instead get an ellipse, a hyperbola, or two parallell lines.

Quadratic forms also work in higher dimensions, but this answer is getting long as it is.
If you want, you can probably read more in a standard linear algebra textbook.

Note: @MarkBennet results gave me an idea of alternative approach.

Lets start off by assuming that $ax^{2}+2hxy+by^{2}=0$ represent a pair of straight lines (real/coincident/imaginary).

Substituting $\frac{y}{x}=m$ we have

$\Rightarrow bm^{2}+2hm+a=0$

$m=\frac{-h \pm \sqrt{h^{2}-ab}}{b}$

Plugging back $m=\frac{y}{x}$ we have two equations,

$$by=(-h + \sqrt{h^{2}-ab})x$$ and $$by=(-h - \sqrt{h^{2}-ab})x$$

These are the two lines represented by our equation $ax^{2}+2hxy+by^{2}=0$ , and as you can see we have an under root expression so we have two cases.

$(i)$ The lines are real and distinct if $h^{2}-ab>0$.

$(ii)$The lines are imaginary if $h^{2}-ab<0$.

• @Mark Bennet has a better way. – Aditya Dev Oct 22 '15 at 15:26