設z=x+yi,則
|z+3-4i|
=|x+3+(y-4)i|
=√[(x+3)²+(y-4)²]
∵|Z+3-4i|=2
∴(x+3)²+(y-4)²=4
若x=2sint-3,則y=2cost+4 （t為參數(shù)）
x-1=2sint-4
|z-1|=√[(x-1)²+y²]
=√[(2sint-4)²+(2cost+4)²]
=√[4sin²t-16sint+16+4cos²t+16cost+16]
=√[36+16(cost-sint)]
=√[36-16√2cos(t+45°)]
因為-1≤cos(t+45°)≤1
所以√(36-16√2)≤|z-1|≤√(36+16√2)
∵36-16√2=(4√2-2)²,36+16√2=(4√2+2)²
∴4√2-2≤|z-1|≤4√2+2若x=2sint-3，則y=2cost+4 （t為參數(shù)）這參數(shù)怎么設的啊（就是怎么來的）圓的標準方程： （x-a)²+(y-b)²=r²所對應的參數(shù)方程：x-a=rsint,y-b=rcostorx=a+rsint,y=b+rcost