已知a,b属于R,圆C1:x^2+y^2-4x+2y-a^2+5=0与圆C2:x^2+y^2-(2b-10)x-2by+2b^2-10b+16=0交于不同两点A(x1已知a,b属于R,圆C1:x^2+y^2-4x+2y-a^2+5=0与圆C2:x^2+y^2-(2b-9)x+3by+2b^2-10b+16=0交于不同两点A(x1,y1)B(x2,y2),且(x1-x2)/(y1-y
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已知a,b属于R,圆C1:x^2+y^2-4x+2y-a^2+5=0与圆C2:x^2+y^2-(2b-10)x-2by+2b^2-10b+16=0交于不同两点A(x1已知a,b属于R,圆C1:x^2+y^2-4x+2y-a^2+5=0与圆C2:x^2+y^2-(2b-9)x+3by+2b^2-10b+16=0交于不同两点A(x1,y1)B(x2,y2),且(x1-x2)/(y1-y
已知a,b属于R,圆C1:x^2+y^2-4x+2y-a^2+5=0与圆C2:x^2+y^2-(2b-10)x-2by+2b^2-10b+16=0交于不同两点A(x1
已知a,b属于R,圆C1:x^2+y^2-4x+2y-a^2+5=0与圆C2:x^2+y^2-(2b-9)x+3by+2b^2-10b+16=0交于不同两点A(x1,y1)B(x2,y2),且(x1-x2)/(y1-y2)+(y1+y2)/(x1+x2)=0,则实数b的值为
已知a,b属于R,圆C1:x^2+y^2-4x+2y-a^2+5=0与圆C2:x^2+y^2-(2b-10)x-2by+2b^2-10b+16=0交于不同两点A(x1已知a,b属于R,圆C1:x^2+y^2-4x+2y-a^2+5=0与圆C2:x^2+y^2-(2b-9)x+3by+2b^2-10b+16=0交于不同两点A(x1,y1)B(x2,y2),且(x1-x2)/(y1-y
答:
依据题意知道:
x1²+y1²-4x1+2y1-a²+5=0……………………………………(1)
x1²+y1²-(2b-10)x1-2by1+2b²-10b+16=0……………………(2)
x2²+y2²-4x2+2y2-a²+5=0………………………………………(3)
x2²+y2²-(2b-10)x2-2by2+2b²-10b+16=0……………………(4)
由(1)减去(3)得:
(x1-x2)(x1+x2)+(y1-y2)(y1+y2)-4(x1-x2)+2(y1-y2)=0………………(5)
由(2)减去(4)得:
(x1-x2)(x1+x2)+(y1-y2)(y1+y2)-(2b-10)(x1-x2)-2b(y1-y2)=0…………(6)
由(5)减去(6)得:
(b-7)*(x1-x2)+(1+b)(y1-y2)=0………………………………………………(7)
因为:(x1-x2)/(y1-y2)+(y1+y2)/(x1+x2)=0
所以:(b+1)/(7-b)+(y1+y2)/(x1+x2)=0……………………………………(8)
由(5)得:
(x1-x2)(x1+x2-4)+(y1-y2)(y1+y2+2)=0……………………………………(9)
由(7)、(8)和(9)得:
(x1-x2)/(y1-y2)=(1+b)/(7-b)=(y1+y2+2)/(4-x1-x2)=-(y1+y2)/(x1+x2)
令y1+y2=n,x1+x2=n,上式化为:
(1+b)/(7-b)=(n+2)/(4-m)=-n/m
解得:b=-9,m=4,n=-2
所以:b=-9