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

Oct 30, 2015

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

${x}^{2} + {y}^{2} - 2 x + 4 y = - 6$

Now we can complete the squares:

$\left({x}^{2} - 2 x + 1\right) - 1 + \left({y}^{2} + 4 y + 4\right) - 4 = - 6$

Which we can rewrite into

${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = - 1$

We know that a circumference with center $\left({x}_{0} , {y}_{0}\right)$ and radius $r$ is represented by the equation

${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$.

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.