Interpreting relationship between points on a quadratic curve

Question

Taking a quadratic $Q(x)=ax^2+bx+c$, we can rearrange it to $$\frac{Q(x)-c}{x}=ax+b$$ Then, supposing there are two points $(x_1,y_1),(x_2,y_2)\in Q$, we have:

$$a=\frac{\frac{y_1-c}{x_1}-\frac{y_2-c}{x_2}}{x_1-x_2}$$

Bringing in a third point, $(x_3, y_3)$, we can take:

$$\frac{\frac{y_1-c}{x_1}-\frac{y_2-c}{x_2}}{x_1-x_2}=\frac{\frac{y_2-c}{x_2}-\frac{y_3-c}{x_3}}{x_2-x_3}$$

and solve for $c$. I have found that:

$$c=\frac{x_1^2(x_2y_3-x_3y_2)+x_2^2(x_3y_1-x_1y_3)+x_3^2(x_1y_2-x_2y_1)}{x_1^2(x_2-x_3)+x_2^2(x_3-x_1)+x_3^2(x_1-x_2)}$$ with both numerator and denominator forming a nice cyclical pattern of terms.

My struggle here is trying to visualise what the numerator and denominator each represent for three points, given that $c$, the y intercept of the only quadratic to go through them, is their ratio.

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TL,DR: Given three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is there a way to interpret, perhaps graphically, the following?

$$\alpha=x_1^2(x_2y_3-x_3y_2)+x_2^2(x_3y_1-x_1y_3)+x_3^2(x_1y_2-x_2y_1)$$ $$\beta=x_1^2(x_2-x_3)+x_2^2(x_3-x_1)+x_3^2(x_1-x_2)$$

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After much consideration, I so far have that for the vectors:

$$\mathbf{X}=(x_1,x_2,x_3), \mathbf{X^2}=(x_1^2,x_2^2,x_3^2), \mathbf{Y}=(y_1,y_2,y_3), \mathbf{1^3}=(1,1,1)$$ then $$c= \frac{\mathbf{Y}\cdot(\mathbf{X^2}\times\mathbf{X})}{\mathbf{1^3}\cdot(\mathbf{X^2}\times\mathbf{X})}$$ altered thanks to triple product cyclicality.

In other words, we can assign a vector $\mathbf{V}=(v_1,v_2,v_3)$ to any set of three $x$ co-ordinates and have that $$c=\frac{y_1v_1+y_2v_2+y_3v_3}{v_1+v_2+v_3} $$ The below image indicates one arrangement, $c$ is the ratio of the dot products between one of the red vectors with the blue vector.

Answer 1

Note that Q(0) = ... = c = the y-intercept.

That means the RHS of your work is a formula for calculating c, the y-intercept.