A ping-pong ball lying inside a wine glass Show that a ping-pong ball with radius $r$, lying inside a wine glass described by the function $x^2$, has its center at $r^2+\frac{1}{4}$ units above the bottom of the glass. Here is a visualization of the problem !1 My best attempt is trying to find the derivative of the circle and the function to find some relationship at the point where they meet. The problem looked very simple at first, but I can't figure it out now.

Solution: Let the circle's equation be $$(x)^2+(y-k)^2=r^2$$ and the Parabola's equation be $$y=x^2$$ Then due to the symmetry of the Parabola both the curves will meet at the points $(a,b)$ and $(-a,b)$. Consequently, $b=a^2$ and $a^2+(b-k)^2=r^2$.Substituting for $a^2$ in the second equation, we get: $$b+(b-k)^2=r^2$$ Moreover, since the Parabola is tangent to the circle, the gradient at these points is the same. This implies that $2a=\frac{a}{k-b}$, which further implies that $k-b=1/2$ as $a\
eq0$. Now, using the two results we can conclude that $b=r^2-1/4$ which implies that $k=r^2+1/4$. The value of $k$ gives you how high the circle is above the x-axis.