方程为 $x^2 + y^2 - 6x + 10y - 27 = 0$ 的圆与直线 $x = -2$ 相交于点 $P$ 和 $Q$。

求由圆在 $P$、$Q$ 两点处的切线以及直线 $x = -2$ 围成的三角形面积。

配方找出圆心，并代 $x=-2$ 求弦的两个端点。两条切线关于经过圆心的水平线对称，因此只需求一条切线与这条水平线的交点。

The circle is $(x-3)^2+(y+5)^2=61$, so its centre is $(3,-5)$. Setting $x=-2$ gives $(y+5)^2=36$, and hence the points $(-2,1)$ and $(-2,-11)$. Their separation is $12$.

The radius to $(-2,1)$ has slope $-6/5$, so the tangent there is $y-1=\frac56(x+2)$. The other tangent is its reflection in $y=-5$, so the two tangents meet on that line. Setting $y=-5$ in the first equation gives $x+2=-\frac{36}5$. This is a horizontal distance of $\frac{36}5$ from the base line $x=-2$. Thus the triangle area is