Question

过点p(3,2)与圆(x-1)^2+y^2=1相切的直线方程

Answer 1

C:(x-1)^2+y^2=1
C(1,0),r=1
y-2=k(x-3)
kx-y+2-3k=0
|k-0+2-3k|/√(1+k^2)=r=1
k=(4±√7)/4
y-2=[(4±√7)/4]*(x-3)