(1)
2mx-y-8m-3=0
2m(x-4)-y-3=0
由題目易知,直線l過一定點(diǎn)P(4,-3)
將定點(diǎn)P(4,-3)代入圓方程左式:x^2+y^2-6x+12y+20中,得
4^2+(-3)^2-6*4+12*(-3)+20 = -15 < 0
說明定點(diǎn)P(4,-3)在圓C內(nèi)部.
所以,不論m為何實(shí)數(shù),直線l與圓總相交.
證畢.
2.將圓方程化為標(biāo)準(zhǔn)形式得:
(x-3)^2 + (y+6)^2 = 5^2
易知,圓心為O(3,-6),半徑為r＝5
要使截得的弦長最短,根據(jù)數(shù)形結(jié)合,易知,當(dāng)點(diǎn)P(4,-3)為相交弦中點(diǎn)時所截得弦長最短.
因弦心距|OP| = √[(3-4)^2+(-6+3)^2] = √10
所以所截得最短弦長為 d = 2√(25-10) = 2√15
而此時弦所在的直線斜率為
k = -1/k' = -1/3
即 2m = -1/3
所以m = -1/6
綜上,知,m = -1/6時,l被圓C截得弦最小,最小值為2√15