x^2+y^2+2x-4y=0x^2+2x+1+y^2-4y+4=5(x+1)^2+(y-2)^2=5当切线斜率不存在时,显然不成立设切线方程:y-2=k(x-4)kx-y-4k+2=0点到直线距离|-k-2-4k+2|/√k^2+1=√525k^2=5k^2+520k^2=5k=±1/2所以x-2y=0-x-2y+2+2=0x+2y-4...