圆x^2 y^2 6x-2y 9的圆心M到直线y=2x的距离
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{2/3x-2/(x+y)[(x+y)/3x-x-y]}/【(x-y)/x】的负二次方={2/3x-[2/(x+y)]*[(x+y)-3x(x+y)]/(3x)}*[x/(x-y)]^2={2/3x-
(x+1)^2+(y-3)^2=6^2(x-2)^2+(y+1)^2=3^2圆心距离5,6-3
高中时学的,给你个思路好了.(X-3)^2+(Y-2)^2=1,表示以(3,2)为圆心,1为半径的圆.令k=Y/X,k表示圆上一点与原点连线的斜率,设此直线为y-kx=0,用点(圆心)到直线(y-kx
x^2+y^2-6x-4y+12=0(x-3)^2+(y-2)^2=1令x-3=cosa,y-2=sinax+y=5+cosa+sina=5+√2sin(a+π/4)x+y最大值5+√2,最小值5-√
令x=2cosx,y=2sinx.令t=(y-4)/(x-4)=(2sinx-4)/(2cosx-4)=(sinx-2)/(cosx-2)所以t(cosx-2)=(sinx-2)sinx-tcosx=
圆C1;x*x+y*y+4x+y+1=0.1圆C2:X*X+Y*Y+2X+2Y+1=0.21式减2式,得2x-y=0,即y=2x.33式代入1式,得5x^2+6x+1=0,得x=-1或x=-1/5,则
令x=cosa则y²=1-cos²a=sin²a所以y=sina所以y-2x=sina-2cosa=√(1²+2²)sin(a-b)=√5sin(a-
(x-2)^2+(y+1)^2=1圆心(2,-1),r1=1(x+1)^2+(y-3)^2=36圆心(-1,3),r2=6圆心距d=√[(2+1)²+(-1-3)²]=5所以d=r
答:x^2+y^2
X1+……+X9~N(0,81)(X1+……+X9)/9~N(0,1)Y/3~N(0,1)(Y1^2+...Y9^2)/9~卡方(9)[(X1+……+X9)/9]/根号[(X1+……+X9)/9/9]
-t是截距的意思,当相切时就是极限点,-t分别可取到最大值和最小值,那么x-y的最值也就知道了再问:极限点是什么意思,,,,点C(3,2)到直线x-y-t=0的距离是什么意思再答:就是取最值的时候,就
1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)(2a+b)(2
我们来一个一个地看.x=2,y=9,z=-4那么x>y=Falsez>x=Falsez>y=False则原题化为FalseorFalseandnotFalse逻辑运算符的运算优先级是not>and>o
令m=x(x+y),n=y(x+y)则m-n=x^2-y^2,于是m-n+1=2x^2,1-m+n=2y^2(m+n-1)^2=4x^2y^2=(m-n+1)(1-m+n)整理得:m^2+n^2-m-
√(x²+y²-2x-2y+2)化为√(x-1)²+(y-1)²就是求圆x²+y²+8y+12=0到(1,1)距离最小和最大.x²
圆的参数方程x=cosθy=sinθ+1(y-1)/(x-2)=k你先画个图,就知道直线y-1=k(x-2)过点(2,1)当p点在圆下和圆相切时的直线,k有最大值此时有圆心(0,1)到直线y-1=k(
根据题意得,9-2m=1,解得m=4.故答案为:4.
解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy
圆1:(x-1)方+y方=1圆心(1,0)半径1圆2:x方+(y-2)方=4圆心(0,2)半径2圆心距=根号3
x^2+y^2-10x-10y=0=(x-5)^2+(y-5)^2=50,为圆心在(5,5)半径为5√2的圆x^2+y^2+6x+2y-40=0为圆心在(-3,-1)半径为5√2的圆设公共弦两端点为C