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数学
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直线Y=x+2与圆x^2+y^2+x-6y+m=0相切,求M值
人气:345 ℃ 时间:2020-10-01 20:04:48
解答
解x^2+y^2+x-6y+m=0(x+1/2)^2+(y-3)^2-1/4-9+m=0(x+1/2)^2+(y-3)^2=9又1/4-m圆心坐标为(-1/2,3)直线方程变为x-y+2=0直线与圆相切,则表示圆心到直线的距离为半径R,则│-1/2-3+2│/√[1² +(-1)²...
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