∴ OA = OB = 2 ∴ AB2 = OA2+OB2 = 22+22 = 8 ∴ AB = 2: OC = AB ∴ OC = 2, which is C (0 0,2).

Another image with parabola y=-x2+mx+n can be solved by point A and point C: the expression of parabola is y=-x2-x+2.

(2)OA = ob∠AOB = 90° ∴∠bao=∠abo=45

∠∠BeO =∠Bao+∠AOE = 45+∠AOE。

∠BeO =∠oef+∠bef = 45 +∠bef∴∠bef=∠aoe

(3) When △EOF is an isosceles triangle, it is discussed in three cases.

① When OE=OF, ∠ ofe = ∠ oef = 45.

In △EOF, ∠ eof =180-∠ oef-∠ ofe =180-45-45 = 90.

∫∠AOB = 90。

Then point e coincides with point a at this time, which does not meet the meaning of the question. This situation does not hold water.

② As shown in Figure ②, when FE=FO,

∠EOF =∠OEF = 45°

In △EOF, ∠ Efo =180-∠ OEF-∠ EOF =180-45-45 = 90.

∴∠ AOF+∠ EFO = 90+90 =180 ∴ ef ∨ ao ∴∠ BEF = ∠ Bao = 45 and ∵ by (.

③ As shown in the answer of Figure ③, when EO=EF, point E is the EH⊥y axis, and point H is in △AOE and △BEF.

∠EAO=∠FBE，EO=EF，∠AOE=∠BEF ∴△AOE≌△BEF∴BE=AO=2

∵eh⊥ob∴∠ehb=90 ∴∠aob=∠ehb ∴eh∥ao∴∠beh=∠bao=45

In Rt△BEH, ∫∠ beh = ∠ ABO = 45 ∴ eh = BH = becos45 = 2× =

∴oh=ob-bh=2-2∴·e(-，2-)

To sum up, when △EOF is an isosceles triangle, the coordinates of point E are E(- 1, 1) or E(-, 2- 2).

(4) p (0,2) or p (- 1, 2)