# Is it a new method to construction of a conic, how can prove?

There are some methods to construct a conic, example: Based on Pascal theorem, Steiner construction, .....I propose a method to construct a conic as follows:

Let $$L_1, L_2$$ be two parallel lines, let $$A, B, C, D$$ be four points in the plane. Let $$E$$ be a point lie on the line $$L_1$$, $$F$$ be the point lie on line $$L_2$$ such that $$EF \parallel AB$$. Let circle $$(E, ED)$$ meets the circle $$(F, FC)$$ at two points $$H$$, $$G$$.

My question: I am looking for a proof that locus of $$H, G$$ is a conic section when $$E$$ be moved on line $$L_1$$.

Without loss of generality, let $$L_1$$ be the x-axis and $$t$$ be a parameter. Denote $$C,D,E,F$$ by $$C(x_1,y_1),D(x_2,y_2),E(t,0),F(t+a,b)$$ such that the vector $$$$ is in the same direction as $$\overrightarrow{AB}$$ and $$x_1,y_1,x_2,y_2,a,b$$ are constants. Then the locus of $$H$$ (or $$G$$) satisfies $$\frac{PE}{ED}=\frac{PF}{FC}=1,$$ where $$P(x,y)$$ represents $$H$$ (or $$G$$). In terms of equations, one has $$(x-t)^2+y^2=(x_2-t)^2+y_2^2\qquad (1)$$ and $$(x-t-a)^2+(y-b)^2=(x_1-t-a)^2+(y_1-b)^2\qquad (2)$$ Then from (1), one gets $$t=\frac{x^2+y^2-x_2^2-y_2^2}{2(x-x_2)}\qquad (3)$$ Subtracting (2) from (1) gives $$(2x-2t-a)(a)+(2y-b)(b)=(x_1+x_2-2t-a)(x_2-x_1+a)+(y_2+y_1-b)(y_2-y_1+b)\quad (4)$$ Now substituting (3) into (4) and clearing denominator, one gets a quadratic equation in $$x$$ and $$y$$, so it gives an equation for a conic.