It is known that the two focal points of ellipse E: X2A2+Y2B2 = 1 (A > B > 0) are F 1, F2, P is on ellipse E, PF 1⊥F 1F2,

(1) because point p is on ellipse c, 2a = | PF 1 |+PF2 | = 10, and a = 5.

At Rt△PF 1F2, | F 1F2 | = | PF2 | 2? |PF 1|2=8，

Therefore, the half focal length of the ellipse c=4,

So b2=a2-c2=3,

So the equation of ellipse C is X225+Y29 = 1.

(2) The equation of a known circle is (x+3)2+(y- 1)2=5.

So the coordinate of the center m is (-3, 1).

Let the coordinates of A and B be (x 1, y 1) and (x2, y2) respectively.

By the questions x 1≠x2 and X1225+Y129 =1,① X2225+Y229 = 1, ②

From ① to ② (x 1? x2)(x 1+x2)25+(y 1？ y2)(y 1+y2)9=0。 ③

Because a and b are symmetrical about point m,

So x 1+x2=-6, y 1+y2=2,

Substitute ③ to get y 1? y2x 1？ x2=2725，

That is, the slope of the straight line L is 2725,

So the equation of the straight line L is y- 1=2725(x+3),

That is 27x-25y+ 106 = 0.