(2) existence
It is known that two points A and B are symmetrical about the symmetry axis, A (- 1, 0), so the symmetry axis x= 1 is the middle perpendicular of the straight line AB, so PA=PB. In triangular PAC, PA-PC
Connect AC, extend the symmetry axis to P, and the distance difference between point P and points B and C is the largest. Let the straight line AC be y=kx+b, then 0=-k+b-3=b∴k=-3 ∴y=-3x-3 so that x= 1, and y=-6∴P( 1,-6). -2x-3 is x? -2x-3-n=0 Let the intersection of the symmetry axis and the X axis be d, let △DMN be an isosceles right triangle according to symmetry, and let M(x 1, y 1) and N(x2, y2), then x2- 1 = n,1. X1x 2 =-3-n = (1√17)/2, so the radius of this circle is (1+√ 17)/2 or = (√17.