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 $(a,b)$.
1. Given that the point \((a, b)\) lies on the circle \(x^2 + y^2 = 1\), we have:
   \[
   a^2 + b^2 = 1
   \]

2. The tangent to the circle at \((a, b)\) has a slope that is the negative reciprocal of the slope of the radius at that point. The slope of the radius is \(\frac{b}{a}\), so the slope of the tangent is:
   \[
   -\frac{a}{b}
   \]

3. The equation of the tangent line at \((a, b)\) can be written in point-slope form:
   \[
   y - b = -\frac{a}{b}(x - a)
   \]
   Simplifying, we get:
   \[
   y = -\frac{a}{b}x + \frac{a^2}{b} + b
   \]
   Using \(a^2 + b^2 = 1\), we can rewrite the equation as:
   \[
   y = -\frac{a}{b}x + \frac{1}{b}
   \]

4. We need to find the intersection of this tangent line with the parabola \(y = x^2 + 1\). Setting the equations equal to each other:
   \[
   x^2 + 1 = -\frac{a}{b}x + \frac{1}{b}
   \]
   Multiplying through by \(b\) to clear the fraction:
   \[
   bx^2 + b = -ax + 1
   \]
   Rearranging terms:
   \[
   bx^2 + ax + (b - 1) = 0
   \]

5. For the tangent to intersect the parabola at exactly one point, the quadratic equation must have exactly one solution. This occurs when the discriminant is zero. The discriminant of \(bx^2 + ax + (b - 1) = 0\) is given by:
   \[
   \Delta = a^2 - 4b(b - 1)
   \]
   Setting the discriminant to zero:
   \[
   a^2 - 4b(b - 1) = 0
   \]
   Substituting \(a^2 = 1 - b^2\) from the circle equation:
   \[
   1 - b^2 - 4b^2 + 4b = 0
   \]
   Simplifying:
   \[
   1 - 5b^2 + 4b = 0
   \]
   This is a quadratic equation in \(b\). Solving for \(b\):
   \[
   5b^2 - 4b - 1 = 0
   \]
   Using the quadratic formula \(b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\) where \(A = 5\), \(B = -4\), and \(C = -1\):
   \[
   b = \frac{4 \pm \sqrt{16 + 20}}{10} = \frac{4 \pm \sqrt{36}}{10} = \frac{4 \pm 6}{10}
   \]
   Thus, we have:
   \[
   b = 1 \quad \text{or} \quad b = -\frac{1}{5}
   \]

6. For \(b = 1\):
   \[
   a^2 + 1^2 = 1 \implies a^2 = 0 \implies a = 0
   \]
   So, one point is \((0, 1)\).

7. For \(b = -\frac{1}{5}\):
   \[
   a^2 + \left(-\frac{1}{5}\right)^2 = 1 \implies a^2 + \frac{1}{25} = 1 \implies a^2 = \frac{24}{25} \implies a = \pm \frac{2\sqrt{6}}{5}
   \]
   So, the other points are \(\left(\frac{2\sqrt{6}}{5}, -\frac{1}{5}\right)\) and \(\left(-\frac{2\sqrt{6}}{5}, -\frac{1}{5}\right)\).

The final answer is \( \boxed{ (0, 1) } \), \(\left(\frac{2\sqrt{6}}{5}, -\frac{1}{5}\right)\), and \(\left(-\frac{2\sqrt{6}}{5}, -\frac{1}{5}\right)\).