Suppose $A(x_1,y_1)$ and $B(x_2,y_2)$ are points on a parabola $y^2=4ax$ through which a circle is drawn with $AB$ as diameter touching the parabola. Find the $|y_1-y_2|$.
I tried to write the equation of the circle with $AB$ as diameter which is:
$$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$
Now,I need to find a condition to show that this circe touches the parabola. But I don't understand how I can do this. I haven't studied calculus yet.
I'm thinking this question could mean two different things
Circle is tangent at either A or B, which means it passes through a total of 3 points the reason for that is that if a circle is tangent to a point on either branch of the parabola $[y>0$ and $y<0]$ then it cannot again intersect any point of the parabola located on that branch. I consider this interpretation highly unlikely to be right since there are infinite values of $|y_1 -y_2|$
Circle is not tangent through A or B but is tangent to a third point C and hence we get to know two things .
With that we consider parametric points of A, B and C as $(at_{1}^2 , 2at_1), (at_2^2 ,2at_2)$ and $(at_3^2,2at_3)$ respectively. $$AC ⊥ BC \\ \frac{2}{t_1 + t_3} . \frac{2}{t_2 + t_3} = -1 \\ t_3^2 + t_3(t_1 +t_2) + t_1t_2 + 4 = 0 \\$$ Since there exists only one value of $t_3$, $$b^2 -4ac = 0 \\ (t_1 + t_2)^2 - 4t_1t_2 -16 =0 \\ (t_1 - t_2)^2 = 4^2 \\ |t_1 -t_2| = 4 \\ \therefore \ \ \ |y_1 - y_2 | = 8a $$