过点p（2,2）向圆x方+y方=1作两条切线pA,pB,则弦AB所在的直线方程是

假设切点的坐标为(x,y),则有 (x-2,y-2)(x,y)=0 => (x-2)x+(y-2)y=0,=> x^2-2x+y^2-2y=0,=>x^2+y^2-2x-2y=0(式1)又因为切点在圆上,所以(x,y)满足x^2+y^2=1带入(式1)得 1-2x-2y=0(AB所在直线方程)所以AB所在直线方程为...哦额~回答哦，蛮可爱~