We can prove this result by invoking the geometrical properties of a parabola, with resort to minimal calculation. Specifically, I will use two well known properties:
Any point on a parabola is equidistant from its focus and directrix. (Follows from the definition of parabola.)
Let $A$ lie on a parabola with focus $F$ and directrix $\ell$. Then the tangent to the parabola at $A$ bisects $\angle FAA'$,where $A'$ is the projection of $A$ on $\ell$. (Well-known, see this page at cut-the-knot for a proof.)
Now suppose $C$ is the center of the circle (the ping-pong ball) and $A$ is the point where it touches the parabola in the 2nd quadrant. Now let $X,Y$ be the projections of $A$ onto $\ell$ (the directrix of the parabola) and the axis of the parabola respectively. Also, let $F$ be the focus of the parabola. From the parabola $y=x^2$, we'll just take the fact that $FF'=\frac12$ and that $O$ bisects $FF'$, and then forget the axes.
Now let $AX=x$, then from property $1$, $AF=x$. Also, the dashed line $t$, the tangent to $\odot(C)$ and the parabola at $A$, bisects $\angle FAX$ (property $2$), and is perpendicular to the radius $AC$, so $AC$ is the external bisector of $\angle FAX$. (Note that this also follows from the optical property of parabola.) Therefore $\angle FAC=\angle PAC=\angle ACF$ (since $AP||CF$)$\implies FC=FA=x$.
Now we are ready to begin the few computations. $FY=YF'-FF'=AX-FF'=x-\frac12$, $CY=CF-FY=x-\left(x-\frac12\right)=\frac12$. Now from Pythagoras' theorem, $$\begin{align*}AY^2=&AF^2-FY^2=AC^2-CY^2\\ \implies & AC^2-CY^2=AF^2-FY^2\\ \implies &r^2-\left(\frac12\right)^2=x^2-\left(x-\frac12\right)^2\\ \implies &r^2=x.\end{align*}$$
Therefore the desired distance $CO=CF+FO=x+\frac14=r^2+\frac14$. $\blacksquare$