已知圆x^2 y^2 6x-8y 25=r^2与x轴相切,求这个圆截y轴所得弦长
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(1)C1:(X+1)^2+(Y+1)^2=10圆心o1(-1,-1)C2:(X-1)^2+(Y+5)^2=50圆心o2(1,-5)O1O2^2=2^2+4^2=20
圆C1:x²+y²-4x+6y=0(x-2)²+(y+3)²=13圆心坐标为(2,-3)圆C2:x²+y²+2x+8y=0(x+1)
已知x2+y2-6x-2y+10=0,则x+y/x-yx²+y²-6x-2y+10=0x²-6x+9+y²-2y+1=0(x-3)²+(y-1)
(x-1)^2+(y-1)^2=1令x-1=sinay-1=cosa则x=1+sina,y=1+cosax^2+y^2=1+2sina+(sina)^2+1+2cosa+(cosa)^2=3+2(si
点(x,y)在圆x²+y²=1上,设x=sinw,y=cosw,则:x+2y=sinw+2cosw则:x+2y的最大值是√5
x2+y2-10x-2y+26=0,x2+y2-10x-2y+25+1=0,(x-5)2+(y-1)2=0,x-5=0或y-1=0,解得x=5,y=1.
设k=y/x则:y=kx,代入(x-2)2+y2=1则:(x-2)^2+k^2x^2=1(k^2+1)x^2-4x+3=0判别式=16-12(k^2+1)>=0k^2再问:(2)求x2+y2的取值范围
X2+Y2+8X+6Y+25=0x²+8x+16+y²+6y+9=0(x+4)²+(y+3)²=0∴x+4=0y+3=0x=-4y=-3X2+4XY+4Y2分之
设x+y=k,代入x2+y2+2x=0x2+(k-x)2+2x=0x2+k2-2kx+x2+2x=02x2-(2k-2)x+k2=0判别式=(2k-2)2-4*2k2>=04k2-8k+4-8k2>=
1)因为直线过定点A(3,0),而3^2-8*3+9=-6
圆心到直线的距离d=(2-1-m)/根号5.直线和圆相离,d>r=1,所以m
C1:(x+1)^2+(y+4)^2=25C2:(x+2)^2+(y-2)^2=10两圆心距为d=√[(-1+2)^2+(-4-2)^2=√37r1=5r2=√10r1-r2
(1)x2+y2-10x-10y=0,①;x2+y2+6x-2y-40=0②;②-①得:2x+y-5=0为公共弦所在直线的方程;(2)弦心距为:|10+5−5|22+12=20,弦长的一半为50−20
已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*
(1)先把圆的方程化成标准形式:(x+1)2+(y-1)2=1从而圆心为(-1,1),半径为1.∵直线y=x+b与圆相切,∴圆心到直线的距离应该等于1.把直线的方程化成x-y+b=0,从而|−1−1+
将x2+y2+174=4x+y,变形得:(x2-4x+4)+(y2-y+14)=0,即(x-2)2+(y-12)2=0,解得:x=2,y=12,则原式=2×122+12=25.
[(x^2+y^2)-(x-y)^2+2y(x-y)]÷4y=1(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)÷4y=1(4xy-2y^2)4y=12x-y=24x/(4x^2-y^2)
解x+y=8两边平方(x+y)²=64即x²+2xy+y²=64∵xy=12∴x²+y²+24=64∴x²+y²=40(x-y)&
(1)圆的方程化为(x-1)2+(y-2)2=8所以圆心为(1,2),半径为22∴d=|1−2+b|2=22∴b=5或-3(2)假设存在.设A(x1,y1),B(x2,y2)∵OA⊥OB,∴y1x1•
根据题意得Y=y1+y2y2=A(1/x^2)y1=B(x-1)(A、B为待定系数)然后就开始联立Y=A(1/x^2)+B(x-1)当X=2时Y=1;x=1时Y=8可得A/4+B=1A-2B=8解得A