15. As shown in Figure 3, the upper semicircle with center at the origin and radius $r$ intersects the $y$-axis and the parabola $y^{2}=x$ at points $P$ and $M$, respectively. The line $PM$ intersects the $x$-axis at $Q$. Find the limit position of point $Q$ as the radius $r$ approaches zero.
$15 .(2,0)$
Suppose the equation of the circle is $x^{2}+y^{2}=x^{2}$, then the equation of the upper semicircle is $u=\sqrt{r^{2}-x^{2}}$
and the parabola is $y^{2}=x$.
The intersection point is $M(\alpha, \beta)$, and let point $Q(x, 0)$.
Then the equation of the line $P M$ is $y-r=\frac{\beta-r}{\alpha}(x-0)$,
The coordinates of point $Q$ are $\left(\frac{\alpha r}{r-\beta}, 0\right), x=\frac{\alpha r}{r-\beta}$
From (1) and (2), we have $x^{2}+x-r^{2}=0$,
The positive root is $\alpha=\frac{-1+\sqrt{1+4 r^{2}}}{2}=\frac{2 r^{2}}{\sqrt{1+4 r^{2}}+1}$
Also, $\beta=\sqrt{\alpha}$, since point $M$ is on the upper semicircle, so $\alpha^{2}=r^{2}-\beta^{2}$.
From (3), we get $x=\frac{a r(r+\beta)}{r^{2}-\beta^{2}}=\frac{r(r+\beta)}{a}$,
$$
\begin{aligned}
\lim _{\rightarrow \rightarrow 0} x & =\lim _{r \rightarrow 0} \frac{r\left(r+\sqrt{\frac{2 r^{2}}{\sqrt{1+4 r^{2}}+1}}\right)}{\frac{2 r^{2}}{\sqrt{1+4 r^{2}}}+1} \\
& =\lim _{r \rightarrow 0}\left[\frac{\sqrt{1+4 r^{2}}+1}{2}+\sqrt{\frac{1}{\left.\sqrt{4 r^{2}+1}+1\right)}}\left(\sqrt{1+4 r^{2}}+1\right)\right] \\
& =1+\sqrt{\frac{1}{2 \cdot 2}} \cdot 2=2
\end{aligned}
$$

When $\tau \rightarrow 0$, the limit position of point $Q$ is $(2,0)$.