A circle of radius $r$ is dropped into the parabola $y=x^{2}$. Find the largest $r$ so the circle will touch the vertex. If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.
 A: Following my comment, you just have to compute for which $r$s
$$ f(x)=\sqrt{r^2-x^2} = r-x^2 = g(x) $$
just at $x=0$. Since:
$$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
the critical radius is obviously $r=\frac{1}{2}$.
A: Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0\ \ \vee\ \ x^2+1-2r=0$$ $$x=0\ \ \vee\ \ x^2=2r-1$$ 
$$x=0\ \ \vee\ \ x^=\pm\sqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows
$$(x=0, y=0), (x=\sqrt{2r-1}, y=2r-1), (x=-\sqrt{2r-1}, y=2r-1)$$
Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(\sqrt{2r-1}, 2r-1)$ & $(-\sqrt{2r-1}, 2r-1)$ 
But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(\sqrt{2r-1}, 2r-1)$ & $(-\sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$\pm \sqrt{2r-1}=0\iff r=\frac{1}{2}$$
& $$2r-1=0\iff r=\frac{1}{2}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Maximum radius of circle, }r_{\text{max}}=\color{blue}{\frac{1}{2}}}}$$
A: Maximum radius is also the curvature at x=0. 
Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2
For Parabola y = x^2,
The value of curvature is 2. Therefore R = 1/2
A: Idea is similar with the accepted answer, yet it is simpler.
The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.
The equation of the parabola is: $y=x^2$.
The two graphs must have a single point of contact at: $(0,0)$.
Substitute $y=x^2$ into the equation of circle:
$$y+(y-r)^2=r^2 \Rightarrow y^2+(1-2r)y=0 \Rightarrow y_1=0; y_2=2r-1.$$
We find:
$$x^2=0 \Rightarrow x_1=0;\\
x^2=2r-1 \le 0 \Rightarrow r\le \frac12.$$
