圆(x+1) ^2+(y-3)^2=10与直线x-y=0交于A,B两点则AB弦长为

来源：学生作业帮助网编辑：作业帮时间：2024/07/07 12:56:22

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圆(x+1) ^2+(y-3)^2=10与直线x-y=0交于A,B两点则AB弦长为
圆(x+1) ^2+(y-3)^2=10与直线x-y=0交于A,B两点则AB弦长为

圆(x+1) ^2+(y-3)^2=10与直线x-y=0交于A,B两点则AB弦长为
圆(x+1) ^2+(y-3)^2=10
∴ 圆心C(-1,3),半径是√10
C到x-y=0的距离d=|-1-3|/√(1+1)=2√2
利用勾股定理
|AB|=2√(R²-d²)=2√(10-8)=2√2
即AB的弦长是2√2.

{4x-3y-10=0 3x-2y=0,{3x-4(x-y)=2 2x-3y=1,{2(x+y)-(x-y)=3 (x+y)-2(x-y)=1 若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值 x-y/7 - x+y/10 =1/2 2(x-y)-5(x+y)=3修改、、、、（x-y)/7-(x+y)/10=1/2 2(x-y)-5(x+y)=3 10X+Y+1/3Z=10Y+X 10X+Y+2/3Z=100X+Y 解方程组:(x+3)(y-2)=(x-3)(y+10),(x+1)(y+3)=(x-2)(y+12) 解方程：（x+3)(y-2)=(x-3)(y+10) (x-1)(y+3)=(x+2)(y+12) { x-y/7-x+y/10=1/2 {2(x-y)+5(x+y)=3 { x-y/7-x+y/10=1/2 {2(x-y)+5(x+y)=3 10/x+y+3/x-y=-5 15/x+y-2/x-y=-1（用换元法解此方程组）解分式方程：10/(x+y)+3/(x-y)=-5,15/(x+y)-2/(x-y)=-1 3/(X-Y)+10/(X+Y)=-5 2/(X-Y)-15/(X+Y)=1 解分式方程组 (3x-y)^2+(3x+y)(3x-y),x=1,y=-2 二元一次方程：2(x+y)-(x-y)=3 (x+y)-2(x-y)=1 (x-y)/(x+y)=3求( 3x-2y-1)/(x+y-5) 2(x+y)-3(x-y)=1 6(x+y)+(x-y)=51 {3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1 变量x,y满足约束条件,x+y>=3,x-y>=-1,2x-y (x-y)^2+(x+y)(x-y) 其中 X =3 Y=-1