# How do you find the standard form given x^2+y^2-6x+4y+14=0?

Aug 8, 2018

Shown below

#### Explanation:

Complete the square on both $x$ and $y$

${x}^{2} - 6 x + 9 + {y}^{2} + 4 y + 4 = - 14 + 9 + 4$

${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = - 1$

$\implies \exists x , y \in \mathbb{C}$

In other words when you try and sketch thus graph on a cartesian plane, nothing shows up, no real values of $x$ and $y$ satisfy this equation