圆x^2+y^2-4x+6y=0截x轴所得的弦与截y轴所得的弦的长度比为?

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圆x^2+y^2-4x+6y=0截x轴所得的弦与截y轴所得的弦的长度比为?
圆x^2+y^2-4x+6y=0截x轴所得的弦与截y轴所得的弦的长度比为?

圆x^2+y^2-4x+6y=0截x轴所得的弦与截y轴所得的弦的长度比为?
令y=0,解得x1=0,x2=4:；再令x=0,解得y1=0,y2=-6.即截x轴所得弦的弦长为4,截y轴所得弦的弦长为6,其比值为2比3.

直线的斜率是k则圆被直线所截弦长为sqrt(1+k^2)·|x1-x2| 将直线方程(y=x-5)代入圆方程得 x^2+(x-5)^2-4x+4(x-5)+6=0 化简得2x

令x=0得
y^2+6y=0
|y2-y1|=|0-6|=6
令y=0
x^2-4x=0
|x2-x1|=|0-4|=4
|y2-y1|/|x2-x1|=6/4=3/2

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