Question 1048633
From the system {{{x^2+(y-6)^2 = 36}}} and {{{ x^2 = 4ky }}} we fast-forward to the point where, after substitution and a little algebra, we get

{{{y^2+(4k-12)y = 0}}}.


Case 1. 

We know beforehand that the two curves intersect at (0,0). To ensure that the above equation has at least one solution, let the discriminant 

{{{b^2-4ac = (4k-12)^2 - 4*1*0 = (4k-12)^2 >= 0}}}

===> {{{4k-12 >= 0}}}  ===> {{{k >= 3}}}.

Now when k = 3, the circle will "fit" snugly on the parabola at the origin, hence intersect only at that point.  

We focus on the instance k > 3.

===> {{{y^2+(4k-12)y = y(y+(4k-12)) = 0}}}  ===> y = 0, -(4k-12).

But {{{y = -(4k-12) < 0}}}.  If k > 3 are plugged into {{{ x^2 = 4ky }}}, we will be solving for {{{ x^2 = -4k(4k-12) < 0 }}}, a contradiction, hence not producing new solutions other than the origin.

Therefore, for k in [3, {{{infinity}}}), the parabola {{{ x^2 = 4ky }}} will intersect the circle only at the origin.


Case 2.  For k in ({{{-infinity}}}, 0]:

If k = 0, {{{ x^2 = 4ky }}} becomes {{{x^2 = 0}}}, or the vertical line x = 0, which is not a parabola, so k = 0 is excluded.  
For k < 0, the parabola opens downward (can accept only negative y values), hence it will automatically intersect the circle only at the origin.



Case 3.  0 < k < 3:

It can be shown that for 3/2 < k < 3, the parabola will intersect the circle at two points of equal level at the LOWER semicircular part.
For k = 3/2, the parabola intersects at the points (-6,6) and (6,6) (the diametrical points!).
For 0 < k < 3/2, the parabola will intersect the circle at two points of equal level at the UPPER semicircular part.

Therefore for Case 3, there are exactly 3 points of intersection, including the origin.


The final answer, therefore, is that k should be in ({{{-infinity}}},0)&#8746;[3, {{{infinity}}}), for the origin to be the ONLY intersection point.