已知圆C:(x-1)2 +(y+ 2)2=9,直线l:2ax-y +2a-1=0 求证:无论a为何实数,直线l与圆总相交
问题描述:
已知圆C:(x-1)2 +(y+ 2)2=9,直线l:2ax-y +2a-1=0 求证:无论a为何实数,直线l与圆总相交
答
圆C:(x-1)2 +(y+ 2)2=9,
直线l:2ax-y +2a-1=0 ,y+2=2ax+2a+1代入(x-1)2 +(y+ 2)2=9:
(x-1)^2+(2ax+2a+1)^2=9
(4a^2+1)x^2+2(4a^2+2a-1)x+(4a^2+4a-7)=0
判别式△=4(4a^2+2a-1)^2-4(4a^2+1)(4a^2+4a-7)
=4[(16a^4+16a^3-4a^2-4a+1) - (16a^4+16a^3-24a^2+4a-7)]
=4(20a^2-8a+8)
=16(5a^2-2a+2)
=16[(根号5a -1/根号5)^2-1/5+2]
=16[[(根号5a -1/根号5)^2+9/5]≥144/5
∴无论a为何实数,直线l与圆总相交