一个圆的方程为 $(x - 3)^2 + y^2 = 18$。方程为 $y = mx + c$ 的直线经过点 $(0, -9)$，并且与该圆相切。

求 $m$ 的两个可能值，并且对每个 $m$ 的值，求切线与圆相切点的坐标。

直线经过 $(0,-9)$，所以截距固定。与圆联立后，令判别式为零，再对每个斜率求重根。

The line is $y=mx-9$. Substituting gives

For tangency,

Thus $m=1$ or $m=-7$. The double root is $x=(6+18m)/(2(1+m^2))$. For $m=1$, it is $x=6$, giving $y=-3$. For $m=-7$, it is $x=-6/5$, giving $y=-3/5$. The respective points are $(6,-3)$ and $(-6/5,-3/5)$.

也可先要求圆心到直线的距离等于半径。找到斜率后，切点就是圆心在切线上的垂足。

For $mx-y-9=0$, the distance from $(3,0)$ is $|3m-9|/\sqrt{1+m^2}$. Equating this to $\sqrt{18}$ and squaring gives

For $m=1$, the perpendicular line through the centre is $y=-(x-3)$. Its intersection with $y=x-9$ is $(6,-3)$. For $m=-7$, the perpendicular is $y=(x-3)/7$. Its intersection with $y=-7x-9$ is $(-6/5,-3/5)$.