Point A is on the Y axis, so the abscissa is 0.

Point a is also on the parabola. If x=0 is brought into the parabolic equation, the ordinate of point A can be solved as c.

That is, A(0, c)

Square OABC, point C is on the X axis, so the coordinate of point C is (c, 0).

B point coordinates (c, c)

The square CDEF has an area of 1 and a side length of 1, that is, CF is 1.

So the coordinate of point F is (c+ 1, 0).

D point coordinates (c, 1)

E point coordinates (c+ 1, 1)

The center of the circle P is on the X-axis, and let the coordinate of P be (p, 0).

A and b are both on the circle, so PA=PB.

The coordinate of point P can be found to be (c/2,0).

Point e is also on the circle, so PA = PE.

Bring in and simplify the coordinates of a, p and e represented by the unknown c, and you can get C 2-C-2 = 0.

Solution: c =-1 or c = 2.

As shown in the figure, c>0, so c=- 1 is discarded, so c=2.

The coordinates of point B are (2,2).

(2)

According to the result of (1), the coordinate of point C is (2,0).

By introducing the parabolic equation, we can solve: b = -3/2.

The parabolic equation is y = x 2/4-3x/2+2.

Let y=0, and the coordinates of the intersection of parabola and X axis are (2,0) and (4,0).

It is known that the coordinate of G point is (4,0).

(The coordinates of G point can also be solved according to the parabolic symmetry axis of x=3 and the coordinates of C point of (2,0). )

The coordinate of point F is (c+ 1 0), that is, (3,0).

M is the midpoint of FG, and the coordinate of m can be obtained as (7/2,0).

The coordinates of point E are (c+ 1, 1), that is, (3, 1).

The equation of the straight line EM can be written as y = -2x +7.

The coordinate of the center of the circle P is (c/2,0), that is, (1 0).

Radius = PA = √5

The equation that can write a circle is: (x-1) 2+y 2 = 5.

The linear EM equation is brought into the cyclic equation: (x- 1) 2+(-2x+7) 2 = 5.

Simplification: x 2-6x+9 = 0

This binary linear equation has only one solution x = 3.

Put x=3 back into the linear equation and get y= 1.

Explain that there is only one intersection point between the straight line EM and the circle, and this intersection point is point e.

That is, the straight line EM and the circle p are tangent to the point e.

The straight line EM is the tangent of the circle p.

(3)

①

The parabolic equation is y = x2/4-3x/2+2 = (x-3) 2/4-1/4.

The symmetry axis of parabola is known as x = 3.

Since the coordinates of A and C are known, the equation of straight line AC can be written as: y = -x+2.

The above two equations can be solved simultaneously, and the coordinate of n point is (3,-1).

The coordinates of point Q are (3, q), Q≦- 1.

△ circumference of △ACQ = AC+AQ+CQ

AC = 2√2 is a constant

Therefore, when AQ+CQ is the minimum, the circumference of △ACQ is the shortest.

The symmetry point of point C about x = 3 is point G (4,0).

CQ = GQ

AQ+CQ = AQ+GQ

Between two points, the straight line segment has the shortest distance.

So the minimum value of AQ+GQ is AG = 2√5.

At this time, point Q is the intersection of straight line AG and parabola symmetry axis x=3.

The equation of straight line AG can be written according to the coordinates of point A and point G: y = -x/2+2.

Bring in x=3 to solve y= 1/2.

So when the coordinate of Q point is (3 1/2), AQ+GQ takes the minimum value of 2√5, that is, AQ+CQ takes the minimum value of 2√5.

So the minimum value of the circumference of △ACQ is 2√2+2√5.

②

The coordinate of point F is (3,0), which is also on the parabola symmetry axis x=3.

FQ=t(t is the length of the line segment, so t≥0)

When t=0, point f and point q coincide.

S△ACQ = S△ACF = 1

When t >; 0, and the coordinates of Q point are (3, t) or (3, -t).

When the coordinate of Q point is (3, t), Q point is in the first quadrant.

S△ACQ = S△AOQ+S△OCQ–S△AOC = 2 * 3/2+2t/2-2 * 2/2 = S

That is s = t+ 1.

When the coordinate of Q point is (3, -t), Q point is in the fourth quadrant.

S△ACQ = S△AOC+S△OCQ–S△AOQ = 2 * 2/2+2t/2-2 * 3/2 = S

That is s = t- 1.

To sum up, the functional relationship between s and t is:

When q is in the first quadrant or on the x axis, S = t+ 1.

S = t- 1, when q is at the fourth pixel limit.