Is there a quadratic equation in $x$ and $y$ that could represent a circle or a pair of straight lines?

Question

I recently came across this very nice question in a textbook:

Find the possible values of the parameter $c$ for which $$4x^2 +(c-1)^2y^2 +2cx+6y=0$$ represents a circle.

The solution was easy enough, but it got me wondering about whether a similar equation involving a parameter could be found that could represent EITHER a circle OR a pair of straight lines in the $(x,y)$ plane depending on how we choose $c$? Can anyone think of such an equation?

Answer 1

$ x^2=ay^2$ represents two lines $ x=\sqrt {a} y $ and $ x=-\sqrt {a} y $ for positive $ a $

Answer 2

Consider

$$c(x^2+y^2)+(1-c)(x+y)^2= 1$$

When $c=1$ this is

$$x^2+y^2=1$$

a circle of radius $1$ centred at the origin.

When $c=0$ this is

$$(x+y)^2=1$$

which is the pair of lines $y=1-x$ and $y=-(1+x)$.

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More generally,

$$(x-a)^2+(y-b)^2=r^2$$

is a circle of radius $r$ centred at $(a,b)$. While

$$(mx+ny+p)(sx+ty+u)=0$$

is a pair of lines if $(m,n,p)$ and $(s,t,u)$ are linearly independent and it is not the case that $m=n=0$ or $s=t=0$.

So the equation

$$c[(x-a)^2+(y-b)^2]+(1-c)(mx+ny+p)(sx+ty+u)=cr^2$$

is a circle for $c=1$ and a pair of lines for $c=0$.

Answer 3

If you allow the parameter not to be a factor I think you can get such an equation with, for example:

$$ (x^c+y^c) ^{3-c} = \alpha > 0 $$

Then $c=1$ will yield a pair of non-intersecting lines, whereas $c=2$ is just a plain circumference. Don't really know what other values of $c$ will look like.

Answer 4

According with the theory of conics and quadratic forms, the only possibility for the equation $$4x^2 +(c-1)^2y^2 +2cx+6y=0$$ represent a circle is that $$(c-1)^2=4\iff c=3\text{ or}-1$$ On the other hand the possibility of $c=0$ in order to have a pair of (real) straight lines, is not good because the equation $$4x^2+y^2+6y=0$$ represents the ellipse $$\left(\frac {x}{1.5}\right)^2+\left(\frac{y+3}{3}\right)^2=1$$ Also the general condition $$4(4(c-1)^2\cdot 0)-(c-1)^2(2c)^2-4\cdot 6^2=-c^2(c-1)^2-4\cdot 36=0$$ cannot gives two real straight lines.