The point $(a, b)$ lies on the circle $x^{2}+y^{2}=1$. The tangent to the circle at this point meets the parabola $y=x^{2}+1$ at exactly one point. Find all such points $(a, b)$.
Answer: $(-1,0),(1,0),(0,1),\left(-\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right),\left(\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right)$.

Since any non-vertical line intersecting the parabola $y=x^{2}+1$ has exactly two intersection points with it, the line mentioned in the problem must be either vertical or a common tangent to the circle and the parabola. The only vertical lines with the required property are the lines $x=1$ and $x=-1$, which meet the circle in the points $(1,0)$ and $(-1,0)$, respectively.

Now, consider a line $y=k x+l$. It touches the circle if and only if the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
y=k x+l
\end{array}\right.
$$

has a unique solution, or equivalently the equation $x^{2}+(k x+l)^{2}=1$ has unique solution, i.e. if and only if

$$
D_{1}=4 k^{2} l^{2}-4\left(1+k^{2}\right)\left(l^{2}-1\right)=4\left(k^{2}-l^{2}+1\right)=0 \text {, }
$$

or $l^{2}-k^{2}=1$. The line is tangent to the parabola if and only if the system

$$
\left\{\begin{array}{l}
y=x^{2}+1 \\
y=k x+l
\end{array}\right.
$$

has a unique solution, or equivalently the equation $x^{2}=k x+l-1$ has unique solution, i.e. if and only if

$$
D_{2}=k^{2}-4(1-l)=k^{2}+4 l-4=0 \text {. }
$$

From the system of equations

$$
\left\{\begin{array}{l}
l^{2}-k^{2}=1 \\
k^{2}+4 l-4=0
\end{array}\right.
$$

we have $l^{2}+4 l-5=0$, which has two solutions $l=1$ and $l=-5$. Hence the last system of equations has the solutions $k=0, l=1$ and $k= \pm 2 \sqrt{6}$, $l=-5$. From (5) we now have $(0,1)$ and $\left( \pm \frac{2 \sqrt{6}}{5},-\frac{1}{5}\right)$ as the possible points of tangency on the circle.