一个圆的方程为 $(x - a)^2 + (y - 3)^2 = 20$。直线 $y = \frac12 x + 6$ 在点 $P$ 处与该圆相切。

(a) 证明 $a$ 的一个可能值为 $4$，并求另一个可能值。

(b) 当 $a = 4$ 时，求该圆在点 $P$ 处的法线方程。

(c) 当 $a = 4$ 时，求与 (b) 中所得法线平行的该圆的两条切线方程。

把切线代入圆方程。相切意味着所得二次方程有且只有一个根。

Substituting $y=x/2+6$ gives

Its discriminant is $(3-2a)^2-5(a^2-11)=-a^2-12a+64$. Setting it to zero gives $(a-4)(a+16)=0$. Hence one value is $4$, as required, and the other is $-16$.

也可用圆心到切线的距离等于半径。直线先写成一般式，再代入圆心。

The centre is $(a,3)$, and the tangent is $x-2y+12=0$. The distance condition is

Thus $|a+6|=10$, giving $a=4$ or $a=-16$.

圆的法线经过圆心，斜率是切线斜率的负倒数。

For $a=4$, the centre is $(4,3)$. The normal has slope $-2$, so its equation is $y-3=-2(x-4)$, or $y=-2x+11$.

所求切线的斜率是 $-2$，过切点的半径斜率就是 $1/2$。先求这条直径的两个端点。

The diameter with slope $1/2$ through $(4,3)$ has equation $y=x/2+1$. Substituting it into the circle gives

The contact points are $(0,1)$ and $(8,5)$. The lines of slope $-2$ through them are $y=-2x+1$ and $y=-2x+21$.

也可以直接用判别式求两条平行切线的截距。斜率固定，只有截距是未知数。

Write a candidate as $y=-2x+c$. Substitution gives

For tangency its discriminant must vanish:

Thus the tangents are $y=-2x+1$ and $y=-2x+21$.

利用半径的方向与长度直接找切点。半径斜率为 $1/2$，横向与纵向位移可设为 $2t$、$t$，再用半径长度求 $t$。

Each required tangent has slope $-2$, so the radius to its point of contact has slope $1/2$. Write the displacement from the centre $(4,3)$ to a point of contact as $(2t,t)$. Since the radius has length $\sqrt{20}$,

The two points of contact are therefore $(4+4,3+2)=(8,5)$ and $(4-4,3-2)=(0,1)$. The tangents of slope $-2$ through these points are

or $y=-2x+21$ and $y=-2x+1$.