Let $a,b,h,g,f,c \in \mathbb{R}$. Then the general equation of a conic given by: $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents the equation of a circle iff \begin{align} \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \leq 0, h = 0, a = b \neq 0. \end{align} For the $\Leftarrow$, it is easy to see as if you plug in the values, it can be easily written as $(x + \frac{g}{a})^2 + (y + \frac{f}{a})^2 = \frac{1}{a^2}(g^2 + f^2 - ac) $ which is a circle. I am having trouble with the $\Rightarrow$ part.
I am using the following definition of a circle:
A circle is a set of points $(x,y) \in \mathbb{R}^2$ which satisfies the equation: $(x- X)^2 + (y - Y)^2 = R^2$ for some $(X,Y) \in \mathbb{R}^2$, $R \in \mathbb{R}_{\geq 0}$.
Using this definition of a circle, how do I show the $\Rightarrow$ part of the above result for the general conic to be a circle?