The tangent at $(a,b)$ is perpendicular to the line through the center of the circle $O$ and $(a,b)$, which has slope $\frac{b}{a}$. So the slope of the tangent is the opposite reciprocal of this, which is $-\frac{a}{b}$. The equation of the tangent is of the form $y=-\frac{a}{b}x+c$ with intersecting $(a,b)$ which leads to $b=-\frac{a}{b}a+c$ $b=-\frac{a^2}{b}+c$ $c=b+\frac{a^2}{b}$ $c=\frac{a^2+b^2}{b}$ Then the equation of the tangent is $y=-\frac{a}{b}x+\frac{a^2+b^2}{b}$ Equating this to the parabola gives $x^2+1=-\frac{a}{b}x+\frac{a^2+b^2}{b}$ $x^2+\frac{a}{b}x+1-\frac{a^2+b^2}{b}=0$ For there to be only one intersection, by the quadratic formula, $\frac{a^2}{b^2} = 4- \frac{4(a^2+b^2)}{b}$ $a^2=4b^2-4a^2b-4b^3$ ... no clue what to do now xD

The equation of tangent will be ax+by-1=0.Also this equation and the equation of parabola has unique solution so eliminate any one variable to get a quadratic in the other,then apply the condition ,discriminant =0 for unique solution.Also use the fact that (a,b) lies on the circle.By this we get 3 points.

The equation of tangent is ax＋by＝1.So,the following simultaneous equations have unique solution. ax＋by＝1---① y＝x＾2＋1---② Case1）b＝0 a=±1,so ①⇔x＝±1 and ②⇔y＝2.Then （a,b）＝（±1,0） Case2）b≠0 ①⇔y＝－a/b・x＋1/b, so ②⇔－a/b・x＋1/b＝x＾2＋1⇔bx＾2＋ax＋（b－1）＝0.Discriminant D＝a＾2－4b（b－1）＝0⇔1－b＾2－4b（b－1）＝0⇔5b＾2－4b－1＝0⇔b＝1,－1/5.Then （a,b）＝（0,1）,（±2/5・6＾1/2,－1/5）. By Case1,2 we conclude that （a,b）＝（0,1）,（±1,0）,（±2/5・6＾1/2,－1/5）.