Parabola and a circle touching at the vertex of the parabola This is a problem I faced during solving my math homework. I asked my math teacher yet she didn't return to me with an answer.I thought I could prove it algebraically but I ended up with too much parameters. So the approach I chose was to try and prove it on a specific parabola. The general problem is- Suppose we have a parabola $$y^2=2px$$ and we want to find a circle that touches the parabola at only one point $ (0,0) $. Also we want that circle to be inside the parabola (so $(0,0)$ will be the only common point of the circle and the parabola).
Is it possible? If it is what is the equation of the circle expressed parameterically?
Thanks!
 A: Say $(x-r)^2 + y^2 = r^2$ is the equation of the circle.
$y^2 = 2px$ is the equation of the parabola.
If you equate, you get $x(x+2(p-r)) = 0$.
So for $r \le p$, $(0,0)$ will be the only common point of the circles and the parabola.
A: Hopefully, this is along the lines of what you are looking for.
Firstly, I took the upper half of the circle and parabola as the curves are symmetric:
$y=\sqrt{2px}$ (the parabola) and
$y=\sqrt{r^2-(x-r)^2}$ (the circle cantered so it passes through $(0,0)$)
This equation for the circle simplifies to
$y=\sqrt{2xr-x^2}$
Setting these equal to each other and simplifying
$\sqrt{2px}=\sqrt{2xr-x^2}$
$2px=2xr-x^2$
We know from your question that $x=0$ is a trivial answer so we may divide out by $x$ giving:
$2p=2r-x$
$x=2r-2p$
We know that if the $x>0$ we have a valid solution for the intersection. However, because you don’t want there to be any intersections (apart from the trivial one) we can determine that $x \le 0$ plugging this in we get:
$2r-2p \le 0$
Which simplifies to:
$r \le p$
That means that any circle that has a radius less than $p$ will not intercept the parabola apart from at $(0,0)$
Parametrically a circle is:
$(r \sin{(t)}, r \cos{(t)})$ for $0 \le t < 2\pi$
