圆O_1:(x-3)^2+(y-2)^2=16与O_2:(x-3-根号3)^2+(y-3)^2=4的位置关系
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/16 08:48:13
![圆O_1:(x-3)^2+(y-2)^2=16与O_2:(x-3-根号3)^2+(y-3)^2=4的位置关系](/uploads/image/z/8711619-51-9.jpg?t=%E5%9C%86O_1%3A%EF%BC%88x-3%EF%BC%89%5E2%2B%28y-2%29%5E2%3D16%E4%B8%8EO_2%3A%28x-3-%E6%A0%B9%E5%8F%B73%29%5E2%2B%28y-3%29%5E2%3D4%E7%9A%84%E4%BD%8D%E7%BD%AE%E5%85%B3%E7%B3%BB)
xS[oA+
,,%{ᦶ`xy"r 4m-,6pBm4&3sOh=|aφ
䰿p
$q}B>;\Ƈt0%ެfQݶm2ɀ?x4Ϧlgo0џ;Q_qRlX:s.v+#ŪwOP7=<{`Nk=dǮ-R`GX-ҋ¹&!k9ɔ=95}%o3ۥͦݯ6Q/] 9/K7-'~6֛pp
cg~XftE},R-ضV%[,N&œD:r+r\{VX\̳*~ *OXIţ[ȪaQA
圆O_1:(x-3)^2+(y-2)^2=16与O_2:(x-3-根号3)^2+(y-3)^2=4的位置关系
圆O_1:(x-3)^2+(y-2)^2=16与O_2:(x-3-根号3)^2+(y-3)^2=4的位置关系
圆O_1:(x-3)^2+(y-2)^2=16与O_2:(x-3-根号3)^2+(y-3)^2=4的位置关系
由题意得:
圆1的圆心是(3,2),半径是4;圆2的圆心是(3+根3,3),半径为2,
两圆心之间的距离的平方是d²=(3-3+根3)²+(2-3)²=4
∴d=2
而半径之差正好为2,所以两圆是内切
第一个图是指两圆外切,第二个图是指两圆内切,就是本题的情况
我觉得应该相邻,不知对不对
圆O_1:(x-3)^2+(y-2)^2=16与O_2:(x-3-???有减号吗?√3)^2+(y-3)^2=4的位置关系
圆O_1:(x-3)^2+(y-2)^2=16与O_2:(x-3-根号3)^2+(y-3)^2=4的位置关系
2(x-y)²-3(x-y)(-y-x)
计算(x-y)^2×(x-y)^3×(y-x)×(y-x)^2
(x+y)/2- x-y/2=6 3(x+y)=4(x-y)
(y-x)^4(x-y)^3+2(x-y)^6(y-x)
(x-y)^3×(y-x)^4+(x-y)^2×(y-x)^5
12(x+y)^3(y-x)/【4(x+y)^2(x-y)】
2分之x+y+3分之x-y;4(x+y)-(x-y)谢谢了,
x(x-y)2-(y-x)3
3x+y=2 -y/x+y=123x+y=2 (-y/x)+y=12
3x(x-2y)+(x+2y)(2x-y)-3y(y-2x)
2(x-y)+3(x+y)2-5(x-y)-8(x+y)2-(x-y)
[2/3x-2/x+y(x+y/3x-x-y)]/x-y/x
化简:{2/3x-2/(x+y)[(x+y)/3x-x-y]}/(x+y)/x
计算2/3x-2/x+y(x+y/3x-x-y)/x-y/x
化简3x-2/(x+y)[(x+y)/3x-x-y]÷(x-y)/x
[(x+y)(x-y)-(x-y)的平方+2y(x-y)]除以(4y)请帮我一下写两道计算题[(x+y)(x-y)-(x-y)的平方+2y(x-y)]除以(4y)[(3x+2y)(3x-2y)-(x+2y)(5x-2y)]除以(4x)
(x-y)(x-y)2(y-x) (x+2y-?)(x-2y+?)