Question

P为直线x-y+4=0上一点,PT为圆C：(x-1)^2+(y+1)^2=2的切线,求切线PT的绝对值的最小值.

Answer 1

设P(x,x+4),

切线|PT|=√[(x-1)^2+(x+5)^2-2]

=√(2x^2+8x+24)

=√[2(x+2)^2+16],

x=-2时它取最小值4.