Locus of centre of circle

Solution 1:

First, I'm assuming that whenever you say an intersection, you're talking about a real intersection. Otherwise, there are always four intersections (counting multiplicity).

Second, since a circle and a parabola are bout quadrics, the number of intersections of these two varieties is at most $4$. Therefore, this question is asking for all the zeros to be real.

We'll start by writing the circle as $(x-h)^2+(y-k)^2=r^2$ or that $$ x^2-2hx+h^2+y^2-2ky+k^2=r^2. $$ Since the circle passes through the origin, we can substitute $(0,0)$ to get that $h^2+k^2=r^2$. Hence, the equation for the circle simplifies to $$ x^2-2hx+y^2-2ky=0. $$

Now, let's look for the other intersection points. These points must satisfy $y=x^2$, so, after substitution, the $x$-coordinates of the intersection points are zeros of $$ x^2-2hx+x^4-2kx^2=0. $$ We already know that $x=0$ is a solution, so we can factor that $x$ out to get $$ x(x^3+(1-2k)x-2h)=0. $$

We focus on the cubic part. This is a depressed cubic with a simple discriminant. The discriminant in this case is $$ -4(1-2k)^3-27(-2h)^2. $$ The discriminant is positive if and only if all the solutions are real and distinct.

Therefore, the condition for all the roots of the cubic to be real and distinct (with $r^2=h^2+k^2$) is that $$ 32k^3-48k^2+24k-4-108h^2>0. $$

Now, we finally want the roots to be distinct. While the roots of the cubic are distinct when the condition above holds, it might be that one of the roots of the cubic is zero, which matches the $x$ that we factored out.

Therefore, we, in addition, need that $h\not=0$ in the cubic so that the cubic has a nonzero constant term, and does not have $0$ as a zero.

Summary: There are three conditions which are necessary and sufficient: