圆的方程为 $x^2 + y^2 - 6x + 2y - 15 = 0$，该圆与 $y$ 轴交于点 $A$ 和 $B$。圆在 $A$ 和 $B$ 处的切线相交于点 $P$。

求 $P$ 的坐标。

先找圆心和两个纵截距。弦的两端关于圆心所在水平线对称，切线交点也在这条水平线上。

Completing the squares gives $(x-3)^2+(y+1)^2=25$, with centre $(3,-1)$. Setting $x=0$ gives $(y+1)^2=16$, so the endpoints are $(0,3)$ and $(0,-5)$.

The radius to $(0,3)$ has gradient $-4/3$, so the tangent there is $y-3=\frac34x$. Reflection in $y=-1$ exchanges the two tangents, so their intersection has $y=-1$. Substituting gives $x=-16/3$. Hence $P=(-16/3,-1)$.

还可以利用切线与半径构成的直角三角形。用对称性确定交点所在的水平线后，只需一个勾股关系。

The centre is $C=(3,-1)$ and the endpoints are $A=(0,3)$ and $B=(0,-5)$. Symmetry gives $P=(u,-1)$. The triangle $PAC$ is right-angled at $A$, so

Expanding gives $-6u+9=41$, so $u=-16/3$. Therefore $P=(-16/3,-1)$.

利用半径与切线垂直，找出两个相似的直角三角形。用半弦长与圆心到弦的距离之比求切线交点到弦的距离。

Completing the squares gives $(x-3)^2+(y+1)^2=25$. Let $C=(3,-1)$ and choose $A=(0,3)$ and $B=(0,-5)$. By symmetry, $P$ lies on $y=-1$. Let $M=(0,-1)$, the midpoint of $AB$. Then $CM=3$ and $AM=4$.

The radius $CA$ slopes downwards to the right, so its perpendicular tangent at $A$ slopes upwards to the right. To reach $y=-1$ from $A$, this tangent must go to the left. Hence $P$ is to the left of $AB$. Also, $CA\perp PA$ and $CM\perp AM$, hence $\angle MAP=\angle ACM$. Thus the right-angled triangles $PMA$ and $AMC$ are similar, giving

Hence $PM=16/3$. Since $P$ is to the left of the $y$-axis, its coordinates are $(-16/3,-1)$.

把圆写成上下两条函数分支，分别用链式法则求两端的切线斜率，再联立切线。不把下半圆误当成正平方根分支。

The circle is $(x-3)^2+(y+1)^2=25$. At $x=0$, the two points are $A=(0,3)$ and $B=(0,-5)$. The upper branch and its derivative are

At $A$ the gradient is $3/4$. The lower branch is $y=-1-\sqrt{25-(x-3)^2}$, with the opposite derivative, so its gradient at $B$ is $-3/4$. The tangents are therefore

Equating them gives $\frac32x=-8$, hence $x=-16/3$ and $y=-1$. Thus $P=(-16/3,-1)$.