求以点C(1,-1)为圆心且与直线x-y+6=0相切的圆的标准方程，求过程！！！
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C(1,-1)为圆心
(x-1)^2 +(y+1)^2 =r^2 (1)
2x+2ydy/dx =0
dy/dx = -y/x = 1
=> x= -y (2)
x-y+6=0 (3)
sub (2) into (3)
2x+6=0
x=3
from (2)
y=-3
(3,-3) into (1)
(x-1)^2 +(y+1)^2 =r^2
r^2 = 4+4 =8
equation of circle
(x-1)^2 +(y+1)^2 =8
(x-1)^2 +(y+1)^2 =r^2 (1)
2x+2ydy/dx =0
dy/dx = -y/x = 1
=> x= -y (2)
x-y+6=0 (3)
sub (2) into (3)
2x+6=0
x=3
from (2)
y=-3
(3,-3) into (1)
(x-1)^2 +(y+1)^2 =r^2
r^2 = 4+4 =8
equation of circle
(x-1)^2 +(y+1)^2 =8
追问
可是它最后得32啊
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