(1) If the center of the circle is E(m, n), its symmetrical point about F( 1, 0) can be changed to (2-m,-n) according to the midpoint coordinate formula; If the radius of circle E is equal to the distance from the center of the circle to the tangent x=4, that is, r=|4-m|, then the equation of circle E is (x-m)? +(y-n)？ =|4-m|？

Substitute the symmetry point (2-m, -n) on the circle into the circle equation as (2-2m)? +(-2n)？ = 16-8|m|+m？ = 16 8m+m？

Simplified to 3m? -8m- 12+4n？ = 8m, two equations are obtained, namely

①3m？ +4n？ = 12 that is, m? /4+n？ /3= 1 is an ellipse with the center at (0,0) and the semi-major axis a=2.

②3m？ - 16m+64/3+4n？ = 100/3 or 9(m-8/3)? / 100+3n？ /25= 1 is an ellipse with its center on the semi-long axis of (8/3,0) a =100/9, because when n=0, m = 10/3+8/3 = 6 or -2/3.

So, the trajectory e equation of the center of gravity c is m? /4+n？ /3= 1 is x? /4+y？ /3= 1

(2) The straight line M passes through F( 1, 0), then

(1) When M∑l, we can know that the linear M equation is x= 1, y = √ 3, that is | AB | = 2 √ 3, and | FP | = 4- 1 = 3, then S △ PAB =/kloc-0.

(2) When M and L are not parallel and perpendicular, let M equation be y = k (x- 1), and if k ≠ 0 is substituted into elliptic E equation, there are

(3+4k？ )x？ -8k？ x+4k？ -12=0 has a, and the intersection of b is △ = 64k 4-48k? + 144-64k^4+ 192k？ = 144(k？ + 1)

x 1.2=(4k？ 6√(k？ + 1))/(3+4k？ ) corresponding to y 1, 2= substituted into y=k(x- 1), and the line segment length |AB| is the base.

Then the distance from the point p (4,0) to the straight line y=k(x- 1) is high, and then the area range -rest *-~