Describe the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
Solution:
Substituting $y=x^2-a$ into $x^2+y^2=a^2$, we get$x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0$. Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).
The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$. We must have $2a-1> 0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$). This only results in a single intersection point in the real coordinate plane. Thus, we see that $a> 1/2$.
My questions:
I don't understand the given solution. Can you explain how this method actually works? How does substituting $y=x^2-a$ into $x^2+y^2=a^2$ and then solving for the roots of that equation tell us where these two curves intersect? Why do the other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$
Best Answer
An equation is satisfied only by points that lie on the curve. When you substitute one equation into the other, it means you are looking at points that satisfy both equations at the same time. Points that satisfy both equations at the same time are exactly the points of intersection!
Now when you get to this step here:
$$x^2(x^2-(2a-1))=0$$
If $x^2 \ne 0$ and $(x^2-(2a-1))\ne 0$ then their product can't be equal to $0$. This means that individually either $x^2=0$ or $x^2-(2a-1)=0$. So what they're doing is solving for $x$ in this second case to get,
$$x=\pm \sqrt{2a-1}$$
Because we want $3$ points of intersection and we already got one from $x=0$, we need that square root to give us $2$ separate solutions, which forces $2a-1>0$.
If $2a-1\le 0$ it will not give us any new solutions, since when $2a-1=0$ it makes $x=\pm \sqrt{0}$ which we already have and when $2a-1<0$ it has no solutions because we can't square root negative numbers here.
If that's still unclear feel free to ask in the comments and I'll try to clarify anything.