Question

求过点(3,3)且与圆x^2+y^2-2y-4=0相切的直线方程为.

Answer 1

C:x^2+y^2-2y=4

C(0,1),r=√5

y-3=k*(x-3)

kx-y+3-3k=0

|0-1+3-3k|/√(1+k^2)=r=√5

k=1.5±0.5√10

y-3=(1.5±0.5√10)*(x-3)