过圆外一点(9,6)做圆(x-2)²+(y-3)²=9的切线方程,

问题描述:

过圆外一点(9,6)做圆(x-2)²+(y-3)²=9的切线方程,

设切线为y=k(x-9)+6圆心(2,3)到切线的距离为半径则有9=|-7k+6-3|²/(1+k²)得:9+9k²=49k²+9-42k40k²-42k=0k(20k-21)=0k=0,21/20因此切线有两条:y=6或y=21/20(x-9)+6,即y=21x/20-69/20...膜拜大神-_-#