圆C:x^2+y^2-6x+12y+20=0,直线L:2mx-y-8m-3=0(m属于实数)m为何值时,直线L被圆C截得的弦长最短?并求出这个最小值
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圆C:x^2+y^2-6x+12y+20=0,直线L:2mx-y-8m-3=0(m属于实数)m为何值时,直线L被圆C截得的弦长最短?并求出这个最小值
圆C:x^2+y^2-6x+12y+20=0,直线L:2mx-y-8m-3=0(m属于实数)
m为何值时,直线L被圆C截得的弦长最短?并求出这个最小值
圆C:x^2+y^2-6x+12y+20=0,直线L:2mx-y-8m-3=0(m属于实数)m为何值时,直线L被圆C截得的弦长最短?并求出这个最小值
圆C可化成(x-3)²+(y+6)²=25,直线可化为m(2x-8)-y-3=0
由直线方程可知,当x=4时,y=-3,所以直线恒过定点(4,-3),该定点在园内,因此直线和圆一定有两个交点,由几何性质可知,当直线2mx-y-8m-3=0垂直于过圆心(3,-6)和(4,-3)两点组成的直线时,弦长最短,此时弦心距为(3,-6)和(4,-3)两点距离,其值√10,所以,弦长=2√(25-10)=2√15
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