The point of intersection of the graph of a quadratic function and a circle

Question

Here is a question I found on the website of International Kangaroo Maths Contest. The question goes like this:

A quadratic function $f(x)=x^2+px+q$ is such that its graph intersects the x and y axes at three different points. The circle through these three points intersects the graph of $f$ at a fourth point. What are the coordinates of this fourth point?

What I tried:

First of all, I noted that the graph of $f(x)=x^2+px+q$ intersects the x and y axes at the points $(0,\ q),\ (\alpha,\ 0)$ and $(\beta,\ 0)$ where $$\alpha = \frac{-p+\sqrt{p^2-4q}}2$$ and $$\beta = \frac{-p-\sqrt{p^2-4q}}2$$

However I couldn't go on from here. Please help me if you know the solution. Thanks for the attention.

Answer 1

By Vieta's formulas, we get $$\alpha+\beta=-\frac p1=-p\tag1$$

Let $(x-c)^2+(y-d)^2=r^2$ be the equation of the circle.

Then, we have $$(x-c)^2+(x^2+px+q-d)^2=r^2,$$ i.e. $$x^4+2px^3+(1+p^2+2q-2d)x^2+(-2c+2pq-2pd)x+c^2+(q-d)^2-r^2=0$$

Let $(X,Y)$ be the fourth point. Then, we get, by Vieta's formulas and $(1)$, $$-\frac{2p}{1}=0+\alpha+\beta+X\iff X=-2p-(\alpha+\beta)=-2p-(-p)=-p$$

It follows that the fourth point is $$\color{red}{(-p,q)}$$

Answer 2

The parabola intersects the $x$ axis at points whose $x$ coordinates are the zeroes of the quadratic polynomial. Then these points can be rendered as $(-p/2+\alpha,0)$ and $(-p/2-\alpha,0)$. The relevant point on the $y$ axis is of course $(0,q)$.

Draw the circle. Since it passes through $(-p/2+\alpha,0)$ and $(-p/2-\alpha,0)$ is must have a mirror line at $x=\-p/2$ which is the perpendicular bisector of the chord. The parabola whose equation also has the form $y=(x+(p/2))^-\alpha^2$ by completing the square ($\alpha$ is the same as the $\alpha$ rendered the previous paragraph) has the mirror line $x=\-p/2$ as well. So, if both the circle and parabola pass through $(0,q)$ as its third given point they must also both pass through the mirror image of that point, thus

$(-p,q)$.