# Is the equation 13x² + 13y² - 26x + 52y = -78  a line, parabola, ellipse, hyperbola, or circle?

Jul 25, 2018

No - it contains no real points.

#### Explanation:

Given:

$13 {x}^{2} + 13 {y}^{2} - 26 x + 52 y = - 78$

First note that all of the coefficients are divisible by $13$, so let's divide the whole equation by $13$ to get:

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

Adding $1 + 4 = 5$ to both sides of the equation and rearranging slightly, this becomes:

${x}^{2} - 2 x + 1 + {y}^{2} + 4 y + 4 = - 1$

That is:

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

Note that this is always false for real values of $x$ and $y$, so this equation describes an empty set of values in the $x$ $y$ plane.

Imaginary circle

#### Explanation:

The given equation:

$13 {x}^{2} + 13 {y}^{2} - 26 x + 52 y = - 78$

$13 \left({x}^{2} + {y}^{2} - 2 x + 4 y\right) = - 78$

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

$\left({x}^{2} - 2 x + 1\right) + \left({y}^{2} + 4 y + 4\right) - 5 = - 6$

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

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

Above equation shows a circle with a radius $r = i$ i.e. an imaginary circle with unit radius.