Common tangent to a circle & parabola. 
I atempted it as: 
Let $(h,k)$ be the point where common tangent occurs to both the curves. So slope at this point must be equall to both curves. Thus for parabola, $$\frac{dy}{dx}= \frac{2}{k}$$ ,at point $(h,k)$. Similarly for circle  $$\frac {dy}{dx}=\frac{(3-h)}{k}$$. Equating $\frac{dy}{dx}$ from both conditions, $$\frac {(3-h)}{k}= \frac{2}{k}$$From here $h=1$. As $(h,k)$ lies both on circle & parabola, it must satisfy both equation. But problem is when I put $h=1$ in circle, I get $k=\sqrt{5}$ or $k=-\sqrt {5}$ but putting in parabola $k=2$ or $k=-2$. How different values of $k$? 

Say that tangent at $(t^2, 2t)$ on parabola is tangent to the circle. So the slope of this tangent is $$2y\frac{dy}{dx}=4\implies\frac{dy}{dx}=\frac{1}{t}$$
Equation of the tangent is $$(y-2t)=\frac{1}{t}(x-t^2) \\ \implies x-ty+t^2=0$$
Since this is the tangent to the circle the distance of this line from center of circle is equal to radius.
$$\frac{3+t^2}{\sqrt{t^2+1}}=3  \\ (3+t^2)^2=9(t^2+1) \\ t^4+6t^2+9=9t^2+9 \\ t^2(t^2-3)=0$$
This gives all the 3 possible tangents $t=0, \sqrt3, -\sqrt3$
Since we want the one on the positive side the equattion of tangent is $$x-\sqrt3 y+3=0$$
