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①当∠ABC＝90°时，如图2，过点B作BE⊥x轴于点E，过点C作CF⊥EB交EB延长线于点F，过点C作CG⊥x轴于点G，则∠AEB＝∠F＝∠ABC＝90°．
∴∠BCF＋∠CBF＝∠ABE＋∠CBF＝90°．∴∠BCF＝∠ABE．∴△BCF∽△ABE． ∴
AEBEAB1＝＝＝． BFCFBC2设AE＝a，则BF＝2a．∵BE＝2，∴CF＝2．
∵OG＝OA＋AE－GE＝3＋a－2＝1＋a，CG＝EF＝2＋2a， ∴B（3＋a，2），C（1＋a，2＋2a）． ∵点B，C在函数y＝
k（x＞0）的图像上，∴2(3＋a)＝(1＋a)(2＋2a)＝k． x解得：a1＝1，a2＝－2（舍去）．
∴k＝42． ················································································································· 9分 ②当∠BAC＝90°时，如图3，过点C作CM⊥x轴于点M，过点B作BN⊥x轴于点N．则∠CMA＝∠CAB＝∠ANB＝90°．
∴∠MCA＋∠CAM＝∠BAN＋∠CAM＝90°．∴∠MCA＝∠BAN． 由（1）知∠B＝45°．∴△ABC是等腰直角三角形．∴AC＝AB． 由①知△MAC∽△NBA．∴△MAC≌△NBA（AAS）．∴AM＝BN＝2． 设CM＝AN＝b，则ON＝3＋b．∴B（3＋b，2），C（3－2，b）． ∵点B，C在函数y＝
k（x＞0）的图像上，∴2(3＋b)＝(3－2)b＝k． x解得：b＝92＋12．∴k＝18＋152． ······························································· 12分 综上所述，k＝42或18＋152．
127．解：（1）令y＝x?2＝0，得x＝4，∴A（4，0）．
2令x＝0，得y＝－2，∴B（0，－2）．···································································· 2分 1∵二次函数y＝?x2?bx?c的图像经过A、B两点，
2九年级数学试卷 第 11页（共 6页）
5???8?4b?c＝0，?b＝，∴?解得?2
c?2．?＝??c＝?2．15∴二次函数的关系式为y＝?x2?x?2． ···························································· 4分
2215令y＝?x2?x?2＝0，解得x＝1或x＝4．∴C（1，0）． ····························· 5分
22（2）∵PD∥x轴，PE∥y轴，∴∠PDE＝∠OAB，∠PED＝∠OBA． ∴△PDE∽△OAB．∴
PDOA4＝＝＝2，∴PD＝2PE． ······································· 7分 PEOB2151设P（m，?m2?m?2），则E（m，m?2）．
22215133∴PD＋PE＝3PE＝3×[(?m2?m?2)－(m?2)]＝?m2?6m＝?(m?2)2?6．
22222∵0＜m＜4，∴当m＝2时，PD＋PE有最大值6． ·············································· 10分 （3）当点M在在直线AB上方时，则点M在△ABC的外接圆上，如图1． ∵△ABC的外接圆O1的圆心在对称轴上，设圆心O1的坐标为（55∴()2?(2?t)2＝(?1)2?t2，解得t＝2．
225，－t）． 2∴圆心O1的坐标为（
55，－2）．∴半径为． 2251∴点M的坐标为（，）． ··············································································· 12分
22yy
图1
图2
九年级数学试卷 第 12页（共 6页）
COBMAxCO2DOBMO1AxO1
当点M在在直线AB下方时，作O1关于AB的对称点O2，如图2． ∵AO1＝O1B＝
5，∴∠O1AB＝∠O1BA．∵O1B∥x轴，∴∠O1BA＝∠OAB． 23，0）． 2∴∠O1AB＝∠OAB，O2在x轴上．∴点O2的坐标为 （
521∴O2D＝1，∴DM＝()2?12＝．
22∴点M的坐标为（
215，?）． ········································································· 14分
2221515，）或（，?）．
2222综上所述，点M的坐标为（
九年级数学试卷 第 13页（共 6
页）