自己做网站制作流程新媒体营销案例分析
已知圆上两点P1，P2，坐标依次为 ( x 1 , y 1 ) , ( x 2 , y 2 ) (x_1,y_1),(x_2,y_2) (x1,y1),(x2,y2)，圆的半径为 r r 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 m i d = ( x 1 + x 2 ) / 2 x_{mid} = (x_1+x_2)/2 xmid=(x1+x2)/2
y m i d = ( y 1 + y 2 ) / 2 y_{mid} = (y_1+y_2)/2 ymid=(y1+y2)/2
P1，P2连线的斜率是
k = ( y 1 − y 2 ) / ( x 1 − x 2 ) k = (y_1-y_2)/(x_1-x_2) k=(y1−y2)/(x1−x2)
P1，P2连线的垂线斜率为
m = − 1 / k m = -1/k m=−1/k
则，圆心所在的直线方程是
y − y m i d = m ∗ ( x − x m i d ) y-y_{mid} = m * (x - x_{mid}) y−ymid=m∗(x−xmid)
圆心 ( x 0 , y 0 ) (x_0,y_0) (x0,y0)同时满足
( x 0 − x 1 ) 2 + ( y 0 − y 1 ) 2 = r 2 (x_0-x_1)^2+(y_0-y_1)^2=r^2 (x0−x1)2+(y0−y1)2=r2和 y 0 − y m i d = m ∗ ( x 0 − x m i d ) y_0-y_{mid} = m * (x_0 - x_{mid}) y0−ymid=m∗(x0−xmid)
或
( x 0 − x 2 ) 2 + ( y 0 − y 2 ) 2 = r 2 (x_0-x_2)^2+(y_0-y_2)^2=r^2 (x0−x2)2+(y0−y2)2=r2和 y 0 − y m i d = m ∗ ( x 0 − x m i d ) y_0-y_{mid} = m * (x_0 - x_{mid}) y0−ymid=m∗(x0−xmid)
将直线方程
y 0 = m ∗ ( x 0 − x m i d ) − y m i d y_0 = m*(x_0 -x_{mid})-y_{mid} y0=m∗(x0−xmid)−ymid
代入圆的公式，
得到
( x 0 − x 1 ) 2 + [ m ∗ ( x 0 − x m i d ) + y m i d − y 1 ] 2 = r 2 (x_0-x_1)^2+[m * (x_0-x_{mid})+y_{mid}-y_1]^2=r^2 (x0−x1)2+[m∗(x0−xmid)+ymid−y1]2=r2
展开，
x 0 2 − 2 x 0 x 1 + x 2 2 + m 2 x 0 2 + 2 m x 0 ∗ ( y m i d − m ∗ x m i d − y 1 ) + ( y m i d − m ∗ x m i d − y 1 ) 2 = r 2 x_0^2-2x_0x_1+x_2^2+m^2x_0^2+2mx_0*(y_{mid}-m*x_{mid}-y_1)+(y_{mid}-m*x_{mid}-y_1)^2=r^2 x02−2x0x1+x22+m2x02+2mx0∗(ymid−m∗xmid−y1)+(ymid−m∗xmid−y1)2=r2
整理，
( 1 + m 2 ) x 0 2 + [ 2 m ( y m i d − m ∗ x m i d − y 1 ) − 2 x 1 ] ∗ x 0 + ( y m i d − m ∗ x m i d − y 1 ) 2 + x 1 2 − r 2 = 0 (1+m^2)x_0^2+[2m(y_{mid}-m*x_{mid}-y_1)-2x_1]*x_0+(y_{mid}-m*x_{mid}-y_1)^2+x_1^2-r^2 = 0 (1+m2)x02+[2m(ymid−m∗xmid−y1)−2x1]∗x0+(ymid−m∗xmid−y1)2+x12−r2=0
令，
A = 1 + m 2 A= 1+m^2 A=1+m2
B = 2 m ( y m i d − m ∗ x m i d − y 1 ) − 2 x 1 B= 2m(y_{mid}-m*x_{mid}-y_1)-2x_1 B=2m(ymid−m∗xmid−y1)−2x1
C = ( y m i d − m ∗ x m i d − y 1 ) 2 + x 1 2 − r 2 C= (y_{mid}-m*x_{mid}-y_1)^2+x_1^2-r^2 C=(ymid−m∗xmid−y1)2+x12−r2
则，
x 0 = − B ± B 2 − 4 A C 2 A x_0=\frac{-B± \sqrt{B^2-4AC}}{2A} x0=2A−B±B2−4AC
y 0 = m ∗ ( x 0 − x m i d ) + y m i d y_0= m*(x_0-x_{mid})+y_{mid} y0=m∗(x0−xmid)+ymid
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