That is A(x+x0)+B(y+y0)+2C=0 1).

If PP' is perpendicular to the straight line L, there is an equation: (y-y0)/(x-x0) = b/a.

Namely: b (x-x0)-a (y-y0) = 0 2)

1)*B-2)*A, eliminated by x: 2ABx0+(B? +A？ )y+(B？ -A？ ) y0+2BC=0, so we get y = k1x0+b1y0+r1.

Similarly, if y is removed, x=k2x0+b2y0+r2.

Substituting x and y into the original curve f (x, y) = 0, the symmetric curve equation obtained is:

F(k2x0+b2y0+r2，k 1x 0+b 1 y0+r 1)= 0。