帮忙一道数学题） 如图6,在平面直角坐标系中,以点A（1,0）为圆心,AO为半径的圆交x轴正半轴于另一点B,点F在⊙A上

⑴连接AF,∵CF是切线,∴AF⊥CF,∴AC=√(AF²+CF²)=3,∴OC=2,∴C(-2,0).⑵∵AO=AF,AE=AE,∠AOE=∠AFE=90°,∴ΔAEO≌ΔAEF(HL),∴∠OAE=∠EAE=1/2∠OAF,∵OB=OF,∴∠B=∠AFB,∴∠OAF=∠AFB+∠B=2∠B,即∠B=1/2∠OAF,∴∠B=∠OAE,∴AE∥BF.⑶∵RTΔCOE∽RTΔCFA,∴CE/CA=CO/CF,CE/3=2/(2√2),CE=3√2/2,∴EF=3√2/2,由ΔAEO≌ΔAEF得：OE=EF=3√2/2,∵AE∥BF,∴OD/OE=OA/OB=1/2,∴OD=3√2,∴B(2,0),D(0,3√2),得直线BD解析式：Y=-3√2/2X+3√2.