已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2)m为何值时,L被圆C截得的弦长最短,求此弦长
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![已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2)m为何值时,L被圆C截得的弦长最短,求此弦长](/uploads/image/z/14331580-52-0.jpg?t=%E5%B7%B2%E7%9F%A52mx-y-8m%2B3%3D0%2C%E5%9C%86c%3Ax%5E2%2By%5E2-6x-12y%2B20%3D0%2C%281%29M%E2%88%88R%2C%E8%AF%81%E6%98%8E%EF%BC%9Al%E4%B8%8E%E5%9C%86c%E6%80%BB%E7%9B%B8%E4%BA%A4%EF%BC%882%E5%B7%B2%E7%9F%A52mx-y-8m%2B3%3D0%2C%E5%9C%86c%3Ax%5E2%2By%5E2-6x-12y%2B20%3D0%2C%281%29M%E2%88%88R%2C%E8%AF%81%E6%98%8E%EF%BC%9Al%E4%B8%8E%E5%9C%86c%E6%80%BB%E7%9B%B8%E4%BA%A4%EF%BC%882%EF%BC%89m%E4%B8%BA%E4%BD%95%E5%80%BC%E6%97%B6%2CL%E8%A2%AB%E5%9C%86C%E6%88%AA%E5%BE%97%E7%9A%84%E5%BC%A6%E9%95%BF%E6%9C%80%E7%9F%AD%2C%E6%B1%82%E6%AD%A4%E5%BC%A6%E9%95%BF)
已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2)m为何值时,L被圆C截得的弦长最短,求此弦长
已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2
已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2)m为何值时,L被圆C截得的弦长最短,求此弦长
已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2已知2mx-y-8m+3=0,圆c:x^2+y^2-6x-12y+20=0,(1)M∈R,证明:l与圆c总相交(2)m为何值时,L被圆C截得的弦长最短,求此弦长
解:
(1)
2mx-y-8m-3=0
2m(x-4)-y-3=0
由题目易知,直线l过一定点P(4,-3)
将定点P(4,-3)代入圆方程左式:x^2+y^2-6x+12y+20中,得
4^2+(-3)^2-6*4+12*(-3)+20 = -15 < 0
说明定点P(4,-3)在圆C内部.
所以,不论m为何实数,直线l与圆总相交.
证毕.
2.将圆方程化为标准形式得:
(x-3)^2 + (y+6)^2 = 5^2
易知,圆心为O(3,-6),半径为r=5
要使截得的弦长最短,根据数形结合,易知,当点P(4,-3)为相交弦中点时所截得弦长最短.
因弦心距|OP| = √[(3-4)^2+(-6+3)^2] = √10
所以所截得最短弦长为 d = 2√(25-10) = 2√15
而此时弦所在的直线斜率为
k = -1/k' = -1/3
即 2m = -1/3
所以m = -1/6
综上,知,m = -1/6时,l被圆C截得弦最小,最小值为2√15
2mx-y-8m-3=0 所以(2x-8)m=(y 3) 所以l必然经过(4,-3)这个点