Cirbola!

$The\quad diagram\quad is\quad symmetric\quad about\quad y-axis\\ \therefore \quad let\quad center\quad be\quad (0,a)\\ \therefore \quad eq.\quad of\quad circle\longrightarrow { \quad x }^{ 2 }\quad +\quad { (y-a) }^{ 2 }\quad =\quad 1\\ solving\quad with\quad the\quad equation\quad of\quad the\quad parabola\quad y={ x }^{ 2 }\\ we\quad get\\ \Longrightarrow { \quad x }^{ 2 }\quad +\quad { ({ x }^{ 2 }-a) }^{ 2 }\quad =\quad 1\\ \Longrightarrow \quad { x }^{ 4 }\quad +\quad (1-2a){ x }^{ 2 }\quad +\quad ({ a }^{ 2 }-1)\quad =\quad 0\\ The\quad parabola\quad and\quad the\quad circle\quad will\quad be\quad tangent\quad \\ to\quad each\quad other\quad when\quad this\quad eq.\quad in{ \quad x }^{ 2 }\quad has\quad real\\ and\quad equal\quad roots\\ \Longrightarrow \quad \quad { (1-2a) }^{ 2 }\quad =\quad 4({ a }^{ 2 }-1)\\ \Longrightarrow \quad a\quad =\quad \frac { 5 }{ 4 } \\ the\quad center\quad of\quad the\quad circle\quad is\quad (0,\frac { 5 }{ 4 } )\\ so\quad a+b+c=9\\$

As Rudresh Tomar notes below, the circle will have an equation of the form x^2 + (y-k)^2 = 1. Differentiating this with respect to x gives: 2 x + 2 (y-k) (dy/dx) = 0. But, at the point of intersection, we must have the derivative wrt x of the circle and the derivative wrt x of the parabola equal. Also, the y-coordinates must be equal. So, if (t,t^2) is the point of intersection, we find 2 t + 2 (t^2-k) (2*t) = 0. Solving for k gives k = 1/2 + t^2. Substituting this value of k into the equation of the circle at point (t, t^2) gives: t^2 + (t^2 -1/2 - t^2) = 1. Thus, t^2 = 3/4 and k = 1/2 + 3/4 = 5/4. Thus, the coordinates of the center are (0, 5/4), so that a+b+c = 0 + 5 + 4 = 9.