Question

自点P(4,4)引圆x²+y²=1的两条切线,切点为A、B,则直线AB的方程为?

Answer 1

PA^2=PB^2=PO^2-r^2=4^2+4^2-1=31

圆P:(x-4)^2+(y-4)^2=PA^2=PB^2=31.(1)

x^2+y^2=1.(2)

(2)-(1):

8x-16+8y-16=-30

AB:4x+4y-1=0