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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.

By Vieta's formulas, we get
$$alpha+beta=-frac p1=-ptag1$$

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 $(1)$,
$$-frac{2p}{1}=0+alpha+beta+Xiff X=-2p-(alpha+beta)=-2p-(-p)=-p$$

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

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

$(-p,q)$.