Question

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

Answer 1

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

Explanation:

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

`(3x-1)^2 >= 0`

`(3y+3)^2 >= 0`

So `(3x-1)^2+(3y+3)^2 >= 0` cannot be equal to `-1`.

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