The trajectory equation of the focus of a parabola passing through a fixed point (1,2) and taking the y-axis as the guide line is RT, great Xia, do me a favor

Let its focus be (m, n), then it can be obtained from the definition of parabola
Root [(x-m) ^ 2 + (y-n) ^ 2] = absolute value of X
Two sides of the square reduction: y ^ 2-2mx-2ny + m ^ 2 + n ^ 2 = 0
This equation is a parabolic equation, and it passes (1,2) points,
Taking (1,2) into the simplification, m ^ 2 + n ^ 2-2m-4n + 4 = 0
This is the trajectory equation of the focus

Given that the parabola takes the x-axis as the guide line and passes through the point m (0,2), the trajectory equation of the focus F of the parabola is? X ^ 2 + (Y-2) ^ 2 = 4

According to the definition of parabola, M is the point on the parabola
The distance from m to f is the same as the distance from m to the guide line
Let f (x, y)
The square of the distance between M and F is X & # 178; + (Y-2) &# 178;
The square of the distance from m to the guide line, that is, the x-axis, is 2 & # 178;
So x & # 178; + (Y-2) &# 178; = 2 & # 178; = 4
This is the required trajectory equation

It is known that the parabola y = x ^ 2 + 2x + B intersects with the X axis at two points AB, and the equation of the circle with ab as the diameter and passing through the point (- 1,2) is obtained

∵ AB is the diameter, the center of the circle is the midpoint of AB, the parabola y = x & # 178; + 2x + B = (x + 1) &# 178; + 1-B, that is, the intersection point C (- 1,0) of the symmetry axis on X axis, the center of the circle C is X & # 178; + 2x + B = 0, known from Weida theorem, X1 + x2 = - 2, x1x2 = B, the solution is (x2-x1) &# 178; = 4-4b, that is, (2R) &# 178; = 4-4b, because (x + 1) &# 17

The trajectory equation of the focus of a parabola passing through a fixed point (1,2) and taking the y-axis as the guide line is RT, great Xia, do me a favor