I mean, out of the condition that a circle actually crosses the parabola. My question is when a circle is "inside" a parabola, would it touch part of the parabola other than just the parabola's vertex at the bottom?
So just the curiosity brought me crazy...but would this be possible?? (There was an image to show but my reputation is not enough since I'm a freshmeat :P)
Thanks xx
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Consider the equations for the circle and parabola. For simplicity, we'll have the circle centred at $(0,0)$, and the parabola tangent to the circle at $(0,-a)$, where $a$ is the radius of the circle.
From this, we have the equation $$ x^2+y^2=a^2 $$ for the circle, and $$ y=bx^2-a $$ for the parabola. Now, if they coincide at any point, then the $x$ and $y$ values must be the same. So, substituting the second equation into the first one, we get $$ x^2+(bx^2-a)^2=a^2 $$ or $$ x^2+b^2x^4-2abx^2=0 $$ Now, at $x=0$, we have the "vertex" point, so we can assume that $x\neq 0$ for other solutions. This gives $$ b^2x^2 = 2ab-1 $$ or $$ x = \pm\frac{\sqrt{2ab-1}}b $$ As such, if $2ab-1>0$, then there are more intersection points than just the vertex, given by $$ \left(\pm\frac{\sqrt{2ab-1}}b,\frac{ab-1}b\right) $$ If $2ab-1=0$, then the parabola follows the circle's trajectory in the vicinity of the vertex, but does not coincide. If $2ab-1<0$, then the "intersection points" occur at imaginary values, and thus do not exist in the reals.
On
We may assume the parabola is $y=x^2$, since by the similarity $x=kx',y=ky'$ this becomes $y'=kx'^2$ (all parabolas are similar), and the problem is the same if the plane is stretched by a magnification factor. Let the circle be centered at $(0,k)$ ($k>0$) and have radius $k$ so that it goes through the origin.
Only the lower half of the circle might intersect the parabola, and this lower half has equation $$y=k-\sqrt{k^2-x^2}.$$ Now it is desired in the question that the circle lie "inside the parabola", which becomes that, for $-k \le x \le k$, we have $$[1] \ \ k-\sqrt{k^2-x^2} \ge x^2.$$ This boils down to $x^2 \ge 2k-1$, where at one step we divided by $x^2$. This shows that we need exactly $k \le 1/2$ in order for the circle to lie above the parabola: If $k \le 1/2$ then $x^2 \ge 2k-1$ holds for all $x$, in particular on $[-k,k]$, while if $k>1/2$ there will be positive $x \in [-k,k]$ for which $x^2<2k-1$, against the circle lying above the parabola.
Now assume that indeed $k \le 1/2$. The condition that the lower semicircle meet the parabola at any point is [1] with equality, call that [1']. Only for $k=1/2$ is there a solution of $x^2=2k-1$ and it is at the origin, while if $k<1/2$ the equation $x^2=2k-1$ has no real solutions, and the circle only meets the parabola at the origin. (Recall the step in which division by $x^2$ made the $0$ drop out as a solution to [1'].)
So in conclusion if a circle lying above the parabola goes through the vertex, then it does not meet the parabola at any other points.
It depends on the circle, and on the parabola. If the parabola is very narrow, or the circle very large, then they'd meet at two points additionally to the vertex of the parabola. If the parabola is wide, or the circle small, they'd meet only at the vertex.