Question

How do you classify `13x² + 13y² - 26x + 52y = -78`?

Answer 1

It has the form of a circumference, but it would have negative radius, so it represents no point in the plane.

Explanation:

First of all, note that we have, coefficients apart, two quadratic terms (`x^2` and `y^2`), and two linear terms (`x` and `y`).

Also, we see that the coefficients of the quadratic terms are equal (they're both `13`). This means that we can divide the whole expression by `13`, obtaining

`x^2+y^2 -2x +4y = -6`

Now we can complete the squares:

`(x^2-2x+1)-1 + (y^2 +4y +4) -4 =-6`

Which we can rewrite into

`(x-1)^2 + (y+2)^2 = -1`

We know that a circumference with center `(x_0,y_0)` and radius `r` is represented by the equation

`(x-x_0)^2 + (y-y_0)^2 = r^2`.

Our equation mimics this one, except for the fact that we have `r^2=-1`, which is obviously impossible using real numbers. In this case, the set of points satisfying the equation is the empty set.