Closer to y than to x

Let $\phi$ be the set of all points inside the region enclosed by the parabola $y = 2-x^2$ in the first quadrant. What is the probability that a point in $\phi$ is closer to the $y$-axis than to the $x$-axis? The answer is in the form $\frac{a\sqrt{b}}{c}$ for where $a$, $b$, $c$ are positive integers, $\gcd(a,c) = 1$, and $b$ is square-free. Input the answer as the sum $a+b+c$.