Circle drops...

Consider a general case of a circle, with center at $O(0, y_0)$ , which touches the parabola at $P(a, a^2)$ and $P' (-a,a^2)$ . The gradient at $P(a,a^2)$ is $\dfrac {dy}{dx} = 2x \big|_{x=a} = 2a$ . Therefore the gradient of $OP$ is $-\dfrac 1{2a}$ . And the equation of $OP$ is given by $\dfrac {y-a^2}{x-a} = - \dfrac 1{2a}$ $\implies y = \dfrac 12 - \dfrac x{2a} + a^2$ . And the $y$ -intercept is given by $y_0 = \dfrac 12 + a^2$ , when $x=0$ . The radius of the circle $OP = r = \sqrt{(y_0-a^2)^2 + a^2} = \sqrt{\dfrac 14 + a^2}$ .

When the circle is only touching the vertex, $a=0$ and $r = \dfrac 12 =0.5$ . This is the largest radius possible because if the radius is larger than $0.5$ , the circle will be touching two points on the parabola away from the vertex. If the radius is smaller than $0.5$ , the circle is still touching the vertex. Therefore the answer is $\boxed{0.5}$ .

The equation of the given parabola is $y=x^2$ . So $\dfrac{dy}{dx}=2x$ . At the vertex of the parabola, $x=0$ . So, at this point $\dfrac{dy}{dx}=0$ . Again $\dfrac{d^2y}{dx^2}=2$ . Therefore the radius of curvature of the parabola at it's vertex is $R=\dfrac{[1+(\dfrac{dy}{dx})^2]^{\dfrac{3}{2}}}{|\dfrac{d^2y}{dx^2}|}=\dfrac{1}{2}=\boxed {0.5}$