Question

已知，圆C:x2+y2-8x+12=0,直线l:ax+y+2a=0. (1)当a为何值时，直线l与圆C相切

已知，圆C:x2+y2-8x+12=0,直线l:ax+y+2a=0. (1)当a为何值时，直线l与圆C相切 （2）当直线l与圆C相交于A,B两点，且|AB|=2根号2时，求直线l的方程。 谢谢
你的答案不对 不好意思

Answer 1

x²-8x+16+y²=4
(x-4)²+y²=4
表示圆心为（4，0）半径为2的圆
根据题意
圆心到直线的距离为半径时相切
｜4a+2a｜/√(a²+1)=2
｜3a｜=√(a²+1)
9a²=a²+1
a²=1/8
a=±√2/4

(2)1/2|AB|=√2，半径为2
根据勾股定理，我们知道圆心到直线的距离为√2
所以有｜4a+2a｜/√(a²+1)=√2
36a²=2a²+2
17a²=1
a=±1/√17
直线为1/√17x+y+2/√17=0即x+√17y+2=0
或-1/√17x+y-2/√17=0即x-√17y+2=0
你说的哪一个不对？
你题目的数据准确吗？

Answer 2

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Answer 3

这个题要自己想呀

Answer 4

我看楼上对的