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Mathematical fixed-point problem
Defined by parabola, the distance from this point to the parabola directrix = the distance from this point to the parabola focus =5.

That is 4+P/2=5. So P=2, parabolic equation y 2 = 4p.

Let the center of the circle c (a 2/4, a) and the radius r.

From the chord length equal to 4 and Pythagorean theorem, we can know that R2 = 2 2+(a 2/4) 2, R2 = 4+a 4/16.

The circular equation is (X-A 2/4) 2+(Y-A) 2 = 4+A 4/ 16.

After deformation, (-x/2+1) a 2-2 ya+(x 2+y 2-4) = 0.

Knowing the arbitrariness of A, the coefficients of the letter A in the polynomial on the left of the above formula are all 0, so

(-x/2+ 1)=0, and 2y=0, and x 2+y 2-4.

X=2, y=0, that is, the fixed-point coordinate is (2,0).