解:(1)∵OC=OB∴∠OCB=∠B∵AB是⊙O的直径∴∠ACB=90 °∴∠A+∠B=90 °∵OD⊥AB∴∠A+∠D=90 °∴∠D=∠B=∠OCB∵∠EOC=∠COD∴△OEC∽△OCD
( 2)∵△OEC∽△OCD∴∴OC2=OE·OD∵OC=2,OE=x∴22=x·OD又∵y=∴y=∴自变量x的取值范围是0<x<2
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