Question

1.过点M(2,4)向圆C:(x-1)平方＋(y+3)平方=1引切线,求切线方程

Answer 1

1.设切线l：kx-y-2k+4=0,圆心(1,-3)到l的距离|-k+7|/√(k^2+1)=1,

∴k=24/7或∞,∴l:24x-7y-20=0,或x=2.

2.设P(x,y),则PA^2=2PB^2,(x+1)^2+y^2=2[(x-1)^2+(y-1)^2],

∴C:x^2+y^2-6x-4y=0.