Find the intersection point For a Parabola   AB and CD are any two parallel chords having slope 1 .
C1 is a circle passing through O A B and C2 is a circle passing through O C D where O is the origin . C1 and C2 intersect at ?
We need to find the points where C1 and C2 intersect. I tried assuming a circle passing through OAB and using parametric from to generate a fourth degree equation. I was unable to proceed further . Can anyone give a hint ? I am in high school
 A: If $A=(t^2,2t)$ then $B=((2-t)^2,2(2-t))$ (I'm setting $a=1$). A generic circle passing through $A$, $B$ and $O$ has equation $x^2+y^2+cx+dy=0$. Substituting there the coordinates of $A$ and $B$ we find two equations, which can be solved for $c$ and $d$:
$$
c = -8 + 2 t - t^2, \quad d = 2 t - t^2.
$$
The intersections of circle and parabola can be found solving the system
$$
\cases{
x^2+y^2+cx+dy=0 \cr
y^2=4x \cr
}
$$
We can eliminate $x$ and insert the expressions for $c$ and $d$ found above, to obtain $$
y^4+\left(8 t-4 t^2-16\right) y^2+\left(32 t-16 t^2\right) y=0.
$$
But we already know three solutions of this equation:
$y=0$, $y=2t$ and $y=4-2t$, hence we
can factor it, to get:
$$
y\ (y-2 t)\ (y - 4 + 2 t)\ (4 + y)=0.
$$
We then obtain a fourth solution: $y=-4$, which is independent of $t$. Hence all circles pass through the point $(4,-4)$ (or, restoring units: $(4a,-4a)$).
A: I found a quicker solution after getting the idea from intelligent pauca.
I also assumed a  = 1.
A $(t^2,2t)$ , B$((2-t)^2,2(2-t))$ are points of intersection of the lines with the parabola. (By equating slopes of two parametric points)
We assume a circle $x^2 + y^2 + cx + dy + c = 0$ passing through A and B . It intersects with the parabola $y^2=4x$ . Eliminating x we get the equation $y^4 + y^2(4c + 16) + 16dy = 0 $.
We observe that the sum of roots of this eqn = 0 . We aldready know three solutions to this equation.
$$ (0) + (2t) + (2(2-t)) + Y = 0$$ The fourth point is independent of t and has y - coordinate -4. So the point is (4,-4).
All circles passing through (0,0) and points of intersection of lines having slope 1 with parabola $y^2 = 4x$ must pass through point (4,-4) . Hence the two circles C1 and C2 must intersect at point (4,-4).
