点 $P$ 和 $Q$ 的坐标分别为 $(1,1)$ 和 $(7,11)$。线段 $PQ$ 是一个圆的直径。

(a) 求该圆的方程。

(b) 求该圆在点 $Q$ 处的切线方程。

(c) 圆上另一个 $x$ 坐标为 7 的点为 $R$。

求点 $Q$ 处切线与点 $R$ 处切线的交点坐标。

直径的中点是圆心，从圆心到任一端点的距离是半径。

The centre is the midpoint $\bigl((1+7)/2,(1+11)/2\bigr)=(4,6)$. The squared radius is $(7-4)^2+(11-6)^2=34$, so the equation is

切线与过切点的半径垂直。先求直径 $PQ$ 的斜率，再取负倒数。

The gradient of $PQ$ is $(11-1)/(7-1)=5/3$. The tangent is perpendicular to this diameter, so its gradient is $-3/5$. Through $Q=(7,11)$, its equation is

也可以把切点附近的圆弧写成函数，直接对上半圆的方程求导。

Near $Q$, the circle is on its upper branch $y=6+\sqrt{34-(x-4)^2}$. Thus

The tangent at $Q$ is $y-11=-\frac35(x-7)$.

先代入 $x=7$ 找另一个圆上点。写出两条切线后联立，求它们的交点。

At $x=7$, the circle equation gives $(y-6)^2=25$. Since $Q=(7,11)$, the other point is $R=(7,1)$. The radius to $R$ has gradient $-5/3$, so the tangent there is $y-1=\frac35(x-7)$. Equating the two expressions for $y$ gives

Thus $x=\frac{46}3$ and $y=6$, so the intersection is $\Big(\frac{46}3,6\Big)$.

$Q$ 和 $R$ 关于直线 $y=6$ 对称，切线也彼此对称。交点必在这条对称轴上，只需代入一条切线。

At $x=7$, the points on the circle have $y=6\pm5$, so $R=(7,1)$. Reflection in $y=6$ swaps $Q$ and $R$ and therefore swaps their tangents. The tangents have different gradients and meet on $y=6$. In the tangent at $Q$ from (b),

Their intersection is $\Big(\frac{46}3,6\Big)$.