Jtdylktuy

While working for some homework problems for circle to select the radius for circles, I encountered with radius 0 and centre at origin i.e., $$x^2+y^2=0$$. When I asked about it to the teacher, he said that it is the equation for point circle and also it denotes the pair of imaginary lines with real intersection point.

My doubt

How come two imaginary lines have real intersection? I mean it's imaginary (pff), it can't be on the paper. How will it have a real intersection point?

$0$ is both an imaginary number and a real number. A complex number is a number of the form $a+bi$ for real $a,b$. A real number is any complex number where $b=0$. An imaginary number is any complex number where $a=0$. Zero fits the criteria for both of these categories.

Really the expression "imaginary" is a kind of PR problem for mathematicians. There is nothing so imaginary about imaginary numbers. Moreover... there is nothing so real about real numbers.

Your teacher's answer is perhaps slightly misleading, but not entirely mistaken.

If we restrict our focus to the real numbers, then the graph of $x^2+y^2=0$ is the subset of $mathbb{R}^2$ given by the point at the origin ${ (0,0) }$.

We can also factorise $x^2+y^2$ as $(x-yi)(x+yi)=0$, which is equal to zero if and only if $x-yi=0$ or $x+yi=0$. If the variables $x$ and $y$ are both real, this means that we must have $x=y=0$, since $x = pm yi$ and $0$ is the only complex number that is both real and imaginary.

However, if $x$ and $y$ are allowed to range over $mathbb{C}$, then $x-yi=0$ and $x+yi=0$ are both 'complex lines'. The word line here is slightly misleading because their dimension is $1$ over $mathbb{C}$, not over $mathbb{R}$.

The reason this seems counterintuitive is because we often view $mathbb{C}$ as a plane—and indeed $mathbb{C}$ is $2$-dimensional over $mathbb{R}$. But then $mathbb{C}^2$ is $4$-dimensional over $mathbb{R}$, and then the equations $x-yi=0$ and $x+yi=0$ describe $2$-dimensional subspaces of a $4$-dimensional space. This is understandably difficult to intuit!

It is worth pointing out, also, that $x^2+y^2=0$ is the union of the complex lines $x-yi=0$ and $x+yi=0$, not their intersection.

Maybe your teacher was referring to the factorization

$$x^2+y^2=(x+iy)(x-iy)=0 iff x=iy quad lor quad x=-iy$$

The conic $x^2+y^2=0$ in $mathbf C^2$ splits into two lines since
$$x^2+y^2=(x+iy)(x-iy)$$
so it's the union of the *isotropic lines $y=ix$ and $y=-ix$. The point $(0,0)$ belongs to both lines.

Note: In complex projective geometry, the points at infinity on those lines, $(1:i:0)$ and $(1:-i:0)$ are called the cyclic points.