点 $A$ 和 $B$ 的坐标分别为 $(4, 3)$ 和 $(8, -5)$。一个半径为 10 的圆经过点 $A$ 和 $B$。

(a) 证明该圆的圆心在直线 $y = \frac{1}{2}x - 4$ 上。

(b) 求该圆的两个可能方程。

圆心到弦的两个端点等距，所以圆心位于弦的垂直平分线上。

The midpoint of $AB$ is $(6,-1)$ and the gradient of $AB$ is $(-5-3)/(8-4)=-2$. The perpendicular bisector therefore has equation

or $y=\frac12x-4$. The centre is on this line because it is equidistant from $A$ and $B$.

也可以先用字母表示圆心，将两个距离的平方相等。平方项相消后便得到一条直线。

Let the centre be $(u,v)$. Equating its squared distances to $A$ and $B$ gives

Expanding and cancelling $u^2+v^2$ gives $8u-16v=64$, so $v=\frac12u-4$.

由 (a) 用一个未知数写出圆心，再令圆心到 $A$ 的距离等于 $10$。两个解对应弦两侧的圆心。

Write the centre as $(u,\frac12u-4)$. Its squared distance to $A=(4,3)$ is

Expanding and dividing by $\frac54$ gives $u^2-12u-28=(u-14)(u+2)=0$. The centres are therefore $(14,3)$ and $(-2,-5)$. The circle equations are

弦的中点和一个端点、圆心组成直角三角形。先求圆心到中点的距离，再按垂直平分线的方向拆成横向和纵向位移。

Let $M=(6,-1)$ be the midpoint of $AB$. Then $AM=\sqrt{2^2+4^2}=2\sqrt5$. The right triangle $AMC$ gives

The direction of the perpendicular bisector is $(2,1)$, of length $\sqrt5$. Thus the two displacements from $M$ are $\pm(8,4)$, giving centres $(14,3)$ and $(-2,-5)$. The equations are $(x-14)^2+(y-3)^2=100$ and $(x+2)^2+(y+5)^2=100$.