<s>[INST] 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.
 [/INST] 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 $$\pm \sqrt{2r-1}=0\iff r=\frac{1}{2}$$
& $$2r-1=0\iff r=\frac{1}{2}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Maximum radius of circle, }r_{\text{max}}=\color{blue}{\frac{1}{2}}}}$$
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