Why do must these 4 distinct points lie on a circle?

Question

I am absolutely stumped by the following question:

Show that 4 distinct points on $y=x^2$ lie on a circle $\iff$ the sum of their x-coordinates is 0.

I'm just trying to develop some intuition for the problem. If we have 2 even "groups" of points the result is fairly straightforward, but the problem implies, for example, that $(1,1), (2,4), (3,9), (-6, 36)$ all lie on the same circle.

My attempts so far have involved trying to generalize relationship of our 4 points to always satisfy $x^2 + y^2 = r^2$, but this resulted in a mess of systems of equations without a clear result. There has to be some very elegant solution to this that I'm not seeing.

Thanks

Answer 1

Take three of the points. They decide a circle $(x-a)^2+(y-b)^2=r^2$. The points also satisfy $y=x^2$. Which is to say, the $x$-coordinates of the three points satisfy $$ (x-a)^2+(x^2-b)^2=r^2\\ x^4+(1-2b)x^2-2ax+a^2+b^2-r^2=0\tag{*} $$ which is a fourth-degree equation.

$\Longrightarrow$: If the fourth point also lies on the circle, then its $x$-coordinate also satisfies $(\text *)$. We thus have four distinct solutions to $(\text *)$, and their sum is given by the (negative of the) third degree coefficient, which we see is $0$.

$\Longleftarrow$: If the sum of the four $x$-coordinates is $0$, then the $x$-coordinate of the fourth point must be the fourth solution to $(\text*)$. Since this fourth point lies on the parabola, its $y$-coordinate must be the square of its $x$-coordinate. The point therefore satisfies the circle equation $(x-a)^2+(y-b)^2=r^2$, and therefore lies on the circle.

Answer 2

Hints:

Perpendicular bisector of line connecting $A(1, 1)$ and $B(2, 4)$ is:

$$y-\frac 52=\frac 13 (x-\frac 32)$$

And that of points $C(3, 9)$ and $D(-6, 36)$ is:

$$y-\frac{45}2=-\frac 13(x-\frac 32)$$

Solving the system of these two equation give the coordinates of center of circle as $O(-30, 13)$

The radius of circle is :

$$r^2=(1+3)^2+(1-13)^2=1105$$

so equation of cicle is:

$$(x+30)^2+(y-13)^2=1105$$

If you plot these in wolfram you can see what question claims.

Answer 3

Here is an approach if you have to only prove that the four given points are concyclic.

Given points are $~A ~(1, 1); ~B ~(2, 4); ~C ~(3, 9); ~D ~(-6, 36)$

We take the lowest point on the graph, which is $A$ (min $y$ coordinate). Checking the slopes of $AB, AC$ and $AD ~(3, 4, -5 ~\text {respectively})$ and similarly checking the slopes of $BC$ and $BD$, we conclude that they do form a convex quadrilateral with $A$ and $C$ being the opposite vertices. So all we need to show is that $~\angle BAD + \angle BCD = 180^\circ$

We can use slopes to find angles between the lines or use vectors and its dot products. Using vectors,

$\vec {AB} = (1, 3); \vec {AD} = (-7, 35)$
$\vec {CB} = (-1, -5), \vec {CD} = (-9, 27)$

Using dot product, we can easily find that,

$\cos (\angle BAD) = \dfrac{7}{\sqrt{65}}$

$\cos (\angle BCD) = - \dfrac{7}{\sqrt{65}}$

That shows $\angle BAD + \angle BCD = 180^\circ$.