Question

求与圆x2+y2-6x+4y=0相切斜率为1的直线方程

Answer 1

斜率为1

x-y+a=0

(x-3)²+(y+2)²=13

圆心(3,-2),半径√13

圆心到切线距离等于半径

|3+2+a|/√(1²+1²)=√13

|a+5|=√26

a=-5±√26

x-y-5+√26=0和x-y-5-√26=0