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∵點P(a,b)在直線x+y+1=0上,∴a+b+1=0,∵a2+b2-2a-2b+2=(a-1)2+(b-1)2,∴a2+b2-2a-2b+2的最小值為點(1,1)到直x+y+1=0的距離,∵d=|1+1+1|2=32=322,∴a2+b2-2a-2b+2 ;的最小值為322.
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