Circle Inside Parabola

A point of tangency $(x, y)$ lies on both the circle and parabola:

$y=x^2$

$x^2+(y-1)^2=r^2 \implies y+(y-1)^2=r^2 \implies y^2-y+1-r^2=0$

Each solution for $y$ corresponds with $2$ coordinates since $y=x^2$ has two solutions (the exception is $y=0$ , but this solution does not meet the requirements and can be ignored). Therefore, there should only be one solution to this quadratic, so the discriminant must be $0$ .

$1-4(1-r^2)=0 \implies 4r^2-3=0 \implies r^2 = \frac{3}{4}$

$\boxed{r=\frac{\sqrt{3}}{2}}$

Concerning large values of $r$ that also create circles intersecting the parabola twice but are not tangent, these do not have one solution for $y$ . They actually have $2$ , but the reason they do not have $4$ intersections is that one solution is negative, and since $y=x^2$ has no negative y-values, it is extraneous. By setting the discriminant to $0$ , we are excluding these circles.