圆(x-2)²+(y-4)²=1的圆心到直线y=2x-1的距离为?(A(0,2)B(0,-2)P在圆O(x-3)²+(y-4)²=1上,求P
A(0,2)B(0,-2)P在圆O(x-3)+(y-4)=1上,求PA+PB的最大值解方程组 x+y=4 x-y=2(x+1)=4(x-2)圆(x-1)+y=4的圆心坐标和半径分别是A(0,2)B(0,-2)P在圆O(x-3)+(y-4)=1上,求PA+PB的最大值设P(x,y),则有(x-3)+(y-4)=1PA+PB=x+(y-2)+x+(y+2)=2x+2y+8=2(x+y+