两圆x^2+y^2=2与x^2+y^2-2x-4y=0的公共弦所在的直线方程
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两圆x^2+y^2=2与x^2+y^2-2x-4y=0的公共弦所在的直线方程
两圆x^2+y^2=2与x^2+y^2-2x-4y=0的公共弦所在的直线方程
两圆x^2+y^2=2与x^2+y^2-2x-4y=0的公共弦所在的直线方程
这个方法很简单啊,你记住啦,只要将x^2和y^2的系数都化为1,然后可以将两式一相减就可得圆的公共弦方程啦!这是一个规律性的结论!
x^2+y^2=2------------------(1)
x^2+y^2-2x-4y=0------------(2)
(1)-(2)得
2x+4y=2
x+2y-1=0为所求的公共弦所在的直线方程
x^2+y^2=2------------------(1)
x^2+y^2-2x-4y=0------------(2)
(1)-(2)得
2x+4y=2
x+2y-1=0为所求的公共弦所在的直线方程
x^2+y^2=2
x^2+y^2-2x-4y=0
下式减上式,有 2x+4y=2
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