点 $(6,1)$ 和 $(-2,7)$ 分别位于一个圆的直径的两端。

(a) 求这个圆的方程。

(b) 这个圆有两条斜率为 $-\frac{1}{2}$ 的切线。

求这些切线与圆相切的点的 $x$ 坐标的精确值。

直径的中点给出圆心。用圆心到直径端点的距离求半径。

The centre is $\Big(\frac{6+(-2)}2,\frac{1+7}2\Big)=(2,4)$. The squared radius is $(6-2)^2+(1-4)^2=25$. Thus the equation is

也可以用直径所对的圆周角为直角。把圆上一点与两个端点连起来，用垂直条件直接建立方程。

For a point $(x,y)$ on the circle, the vectors to the two ends of the diameter are perpendicular by the angle-in-a-semicircle theorem. Their dot product gives

The same equation includes the endpoints. Expanding and completing the squares gives $(x-2)^2+(y-4)^2=25$.

切线的斜率是 $-\frac12$，所以垂直的半径所在直线的斜率是 $2$。两个切点在这条过圆心的直线上。

The line through the centre perpendicular to the tangents is $y-4=2(x-2)$. Substituting into the circle equation gives

The two required coordinates are $x=2-\sqrt5$ and $x=2+\sqrt5$.

也可以把直线的截距暂时保留为未知数。相切时，直线和圆联立所得的二次方程只有一个根。

Substitute $y=-\frac12x+c$ into the circle:

Tangency requires zero discriminant:

Hence $c=5\pm\frac52\sqrt5$. The double root is $x=\frac{2c}{5}$, so $x=2\pm\sqrt5$.

也可以把圆的上、下半部分分别写成根式并求导。令斜率等于 $-\frac12$ 后，开方产生的符号条件能判断每个切点属于哪一半圆。

For the upper semicircle,

Setting the derivative equal to $-\frac12$ gives

Thus $x>2$. Squaring gives $25-(x-2)^2=4(x-2)^2$, so $(x-2)^2=5$ and $x=2+\sqrt5$.

For the lower semicircle,

The required slope now gives $\sqrt{25-(x-2)^2}=-2(x-2)$, so $x<2$. Squaring gives the same equation and hence $x=2-\sqrt5$.

Both values satisfy their respective unsquared equations. Therefore the two required $x$-coordinates are $2-\sqrt5$ and $2+\sqrt5$.