利用圆上两点和圆半径求解圆心坐标

已知圆上两点P1，P2，坐标依次为$(x_1,y_1),(x_2,y_2)$，圆的半径为$r$，求圆心的坐标。

假定P1，P2为任意两点，则两点连成线段的中点坐标是
$$x_{mid}=(x_1+x_2)/2$$
$$y_{mid}=(y_1+y_2)/2$$
P1，P2连线的斜率是
$$k=(y_1-y_2)/(x_1-x_2)$$
P1，P2连线的垂线斜率为
$$m=-1/k$$
则，圆心所在的直线方程是
$$y-y_{mid}=m\ast (x-x_{mid})$$

圆心$(x_0,y_0)$同时满足
$(x_0-x_1{)}^2+(y_0-y_1{)}^2=r^2$和$y_0-y_{mid}=m\ast (x_0-x_{mid})$
或
$(x_0-x_2{)}^2+(y_0-y_2{)}^2=r^2$和$y_0-y_{mid}=m\ast (x_0-x_{mid})$

将直线方程
$$y_0=m\ast (x_0-x_{mid})-y_{mid}$$
代入圆的公式，
得到
$$(x_0-x_1{)}^2+[m\ast (x_0-x_{mid})+y_{mid}-y_1{]}^2=r^2$$

展开，
$$x_0^2-2x_0{x}_1+x_2^2+m^2{x}_0^2+2m{x}_0\ast (y_{mid}-m\ast x_{mid}-y_1)+(y_{mid}-m\ast x_{mid}-y_1{)}^2=r^2$$

整理，
$$(1+m^2)x_0^2+[2m(y_{mid}-m\ast x_{mid}-y_1)-2x_1]\ast x_0+(y_{mid}-m\ast x_{mid}-y_1{)}^2+x_1^2-r^2=0$$

令，
$A=1+m^2$
$B=2m(y_{mid}-m\ast x_{mid}-y_1)-2x_1$
$C=(y_{mid}-m\ast x_{mid}-y_1{)}^2+x_1^2-r^2$

则，
$$x_0=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$$
$$y_0=m\ast (x_0-x_{mid})+y_{mid}$$

x_1 = 2
y_1 = 4
x_2 = 4
y_2 = 2
r = 2
if (x_1 - x_2 == 0):print('横坐标相同，求解可能出错')exit()
else:x_mid = (x_1 + x_2) / 2y_mid = (y_1 + y_2) / 2k = (y_1-y_2)/(x_1-x_2)m = -1/kA = 1 + m**2B = 2 * m *(y_mid - m * x_mid - y_1)- 2 * x_1C = (y_mid - m * x_mid - y_1)**2 + x_1**2 - r**2print(A, B, C)x_c1 = (-B + ((B**2-4*A*C)**0.5))/(2*A)x_c2 = (-B - ((B**2-4*A*C)**0.5))/(2*A)y_c1 = m * (x_c1 - x_mid) + y_midy_c2 = m * (x_c2 - x_mid) + y_midprint('圆心坐标：',(x_c1,y_c1))print('圆心坐标：',(x_c2,y_c2))

运行结果：

InsCode