# How do you find the center and radius of the circle (3x-1)^2 + (3y+3)^2 = -1?

Jun 11, 2016

This equation has no solutions in Real coordinates $\left(x , y\right)$ so does not define a circle.

#### Explanation:

Note that if $x$ and $y$ are Real numbers then so are $3 x - 1$ and $3 y + 3$, hence:

${\left(3 x - 1\right)}^{2} \ge 0$

${\left(3 y + 3\right)}^{2} \ge 0$

So ${\left(3 x - 1\right)}^{2} + {\left(3 y + 3\right)}^{2} \ge 0$ cannot be equal to $- 1$.

Since the equation has no solution, it defines an empty set of points, not a circle.