圆x^2+y^2-2x+4y-20=0截直线5x-12y+c=0所得弦长为8,则c的值?
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圆x^2+y^2-2x+4y-20=0截直线5x-12y+c=0所得弦长为8,则c的值?
圆x^2+y^2-2x+4y-20=0截直线5x-12y+c=0所得弦长为8,则c的值?
圆x^2+y^2-2x+4y-20=0截直线5x-12y+c=0所得弦长为8,则c的值?
(x-1)^2 (y 2)^2=25所以圆心(1,-2),R^2=25圆心到5X-12Y c=0的距离为d=|5*1 12*2 c|/13=|29 c|/13故有:4^2 d^2=R^2解得:c=10或-68
x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2
x+y/2+x-y/3=6,4(x+y)-3(x-y)=-20
{(x+y)/2+(x-y)/3=6 4(x+y)-3(x-y)=-20
4(x+y)-3(x-y)=-20,2/x+y+3/x-y=6
x*x+2x+y*y-4y+5=0 x= y=
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