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 $k$? 
 A: 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 $$x-\sqrt3 y+3=0$$
A: The points at which a common tangent intersects each of the curves need not be the same. In fact, this circle and parabola cross each other, so they do not have the same tangent at any of their four intersections.  
Since you’re given four lines that are potential solutions, instead of deriving a solution you can proceed by a process of elimination. Note first that the third line is the reflection of the second in the $x$-axis and that both the circle and parabola are symmetric with respect to this axis, so these lines both pass or fail together: there’s no need to check both of them.  
The condition for the line $y=mx+b$ to be tangent to the parabola $y^2=4ax$ is that $b=a/m$. In this case, $a=1$, so this becomes simply $b=1/m$. This eliminates the first of the candidates.  
Next, for a line to be tangent to a circle, its distance from the center of the circle must be equal to the circle’s radius. The center of the circle in this problem is $(3,0)$ and checking the two remaining cases (remember that the third line is symmetric to the second) eliminates choice D, leaving B and C.  
If you want to work out the common tangents directly, there are a few ways to proceed. This other answer gives one approach that uses derivatives. Another is to start with the generic line $y=mx+b$ (none of the choices are of the form $x=k$ so for this problem there’s no need to bother checking vertical tangents, which can’t be expressed in slope-intercept form) and work out the conditions for it to be simultaneously tangent to the two curves. If you didn’t happen know the $b=a/m$ condition for the parabola, it’s easy enough to derive. Substitute for $y$ in the equation of the parabola and compute the discriminant of the resulting quadratic equation in $x$: it is $16(1-bm)$. We want a single intersection point, therefore $bm=1$. The discriminant of the quadratic that results from substitution into the circle’s equation is $-4(b^2+6bm-9)$. Again, this must vanish for a single intersection point, and setting $bm=1$ produces $b^2=3$, so the common tangents are $y=\pm\left(\frac1{\sqrt3}x+\sqrt3\right)$.
