(Ⅱ)设直线l的方程为y=kx+2,将直线方程代入圆方程得x2+(kx+2)2-12x+32=0,
整理得(1+k2)x2+4(k-3)x+36=0. ①
直线与圆交于两个不同的点A,B等价于△=[4(k-3)]2-4×36(1+k2)=42(-8k2-6k)>0,
解得-
3 |
4 |
3 |
4 |
(Ⅲ)设A(x1,y1),B(x2,y2),则
OA |
OB |
x1+x2=-
4(k-3) |
1+k2 |
PQ |
所以,
OA |
OB |
PQ |
3 |
4 |
由(Ⅱ)知k∈(-
3 |
4 |
OA |
OB |
PQ |
3 |
4 |
3 |
4 |
OA |
OB |
4(k-3) |
1+k2 |
PQ |
OA |
OB |
PQ |
3 |
4 |
3 |
4 |