Area of all the Circles

If $P(x_{0}, x^{2}_{0}-6)$ is the center point, then the family of circles in question has the requirement $x_{0} = x^{2}_{0}-6$ to ensure tangency to both coordinate axes. This yields $x^{2}_{0} - x_{0} - 6 = (x_{0}-3)(x_{0}+2) = 0 \Rightarrow x_{0} = -2, 3$ . Since the parabola is symmetric with respect to the $y-$ axis, we have four such circles with center points $(\pm3,3); (\pm2,-2)$ and corresponding radii of $3$ and $2$ . The total circular area calculates to $2(\pi2^2 + \pi3^2) = \boxed{26\pi}.$