maoyaoyu
幼苗
共回答了13个问题采纳率:84.6% 举报
利用几何作图法可省掉许多计算.
![](https://img.yulucn.com/upload/c/ba/cbac9a84677c15c7c99330b63c1b726b_thumb.jpg)
已知A1(x1,y1),A2(x2,y2),其中点A0(x0,y0)=(x1+x2)/2,√y0=(y1+y2)/2;圆心M(a,b)待求;
直线A1A2的斜率k1=(y2-y1)/(x2-x1),该线段是圆M的弦,MA1=MA2=r,
直线MA0垂直于A1A2,其斜率为k=(x1-x2)/(y2-y1),另有MA0=√[r^2-(X2-X1)^2/4-(Y2-Y1)^2/4];
a=(x1+x2)/2±MA0*(y2-y1)/[(x2-x1)^2+(y2-y1)^2];
b=(y1+y2)/2±MA0*(x1-x2)/[(x2-x1)^2+(y2-y1)^2];
为简洁,也可将A1A2距离用d代替,则:
a=(x1+x2)/2±√[r^2-d^2/4]*(y2-y1)/d^2;
b=(y1+y2)/2±√[r^2-d^2/4]*(x1-x2)/d^2;
1年前
1