blocks|key|1286988|text|这可能是一个相当深入的计算。这里有一个简单的步骤：http://paulbourke.net/geometry/circlesphere/。一旦你有了圆的方程，你可以简单地把它放在一个包含H和K的形式中。点(h，k)将是圆心。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1286989|(在链接中向下滚动到方程式)|1286990|entityMap|0|LINK|mutability|MUTABLE|url|http://paulbourke.net/geometry/circlesphere/^0|P|18|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

It can be a rather in depth calculation. There is a simple step-by-step here: <a href="http://paulbourke.net/geometry/circlesphere/" rel="noreferrer">http://paulbourke.net/geometry/circlesphere/</a>. Once you have the equation of the circle, you can simply put it in a form involving H and K. The point (h,k) will be the center.

(scroll down a little ways at the link to get to the equations)

blocks|key|3535552|text|这是我的Java端口，用一个非常优雅的IllegalArgumentException避免了行列式消失时的错误条件，我的方法是处理“点是两个相隔很远的”或“点在一条线上”的条件。此外，这将计算半径(并处理异常情况)，这是您的相交坡度方法所不能做到的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3535553|public+class+CircleThree
{+
++static+final+double+TOL+=+0.0000001;

++public+static+Circle+circleFromPoints(final+Point+p1,+final+Point+p2,+final+Point+p3)
++{
++++final+double+offset+=+Math.pow(p2.x,2)+%2B+Math.pow(p2.y,2);
++++final+double+bc+=+++(+Math.pow(p1.x,2)+%2B+Math.pow(p1.y,2)+-+offset+)/2.0;
++++final+double+cd+=+++(offset+-+Math.pow(p3.x,+2)+-+Math.pow(p3.y,+2))/2.0;
++++final+double+det+=++(p1.x+-+p2.x)+*+(p2.y+-+p3.y)+-+(p2.x+-+p3.x)*+(p1.y+-+p2.y);+

++++if+(Math.abs(det)+<+TOL)+{+throw+new+IllegalArgumentException("Yeah,+lazy.");+}

++++final+double+idet+=+1/det;

++++final+double+centerx+=++(bc+*+(p2.y+-+p3.y)+-+cd+*+(p1.y+-+p2.y))+*+idet;
++++final+double+centery+=++(cd+*+(p1.x+-+p2.x)+-+bc+*+(p2.x+-+p3.x))+*+idet;
++++final+double+radius+=+
+++++++Math.sqrt(+Math.pow(p2.x+-+centerx,2)+%2B+Math.pow(p2.y-centery,2));

++++return+new+Circle(new+Point(centerx,centery),radius);
++}

++static+class+Circle
++{
++++final+Point+center;
++++final+double+radius;
++++public+Circle(Point+center,+double+radius)
++++{
++++++this.center+=+center;+this.radius+=+radius;
++++}
++++@Override+
++++public+String+toString()
++++{
++++++return+new+StringBuilder().append("Center=+").append(center).append(",+r=").append(radius).toString();
++++}
++}

++static+class+Point
++{
++++final+double+x,y;

++++public+Point(double+x,+double+y)
++++{
++++++this.x+=+x;+this.y+=+y;
++++}
++++@Override
++++public+String+toString()
++++{
++++++return+"("%2Bx%2B","%2By%2B")";
++++}

++}

++public+static+void+main(String[]+args)
++{
++++Point+p1+=+new+Point(0.0,1.0);
++++Point+p2+=+new+Point(1.0,0.0);
++++Point+p3+=+new+Point(2.0,1.0);
++++Circle+c+=+circleFromPoints(p1,+p2,+p3);
++++System.out.println(c);
++}

}|code-block|syntax|javascript|3535554|请参阅algorithm+from+here|3535555|void+circle_vvv(circle+*c)
{
++++c->center.w+=+1.0;
++++vertex+*v1+=+(vertex+*)c->c.p1;
++++vertex+*v2+=+(vertex+*)c->c.p2;
++++vertex+*v3+=+(vertex+*)c->c.p3;
++++float+bx+=+v1->xw;+float+by+=+v1->yw;
++++float+cx+=+v2->xw;+float+cy+=+v2->yw;
++++float+dx+=+v3->xw;+float+dy+=+v3->yw;
++++float+temp+=+cx*cx%2Bcy*cy;
++++float+bc+=+(bx*bx+%2B+by*by+-+temp)/2.0;
++++float+cd+=+(temp+-+dx*dx+-+dy*dy)/2.0;
++++float+det+=+(bx-cx)*(cy-dy)-(cx-dx)*(by-cy);
++++if+(fabs(det)+<+1.0e-6)+{
++++++++c->center.xw+=+c->center.yw+=+1.0;
++++++++c->center.w+=+0.0;
++++++++c->v1+=+*v1;
++++++++c->v2+=+*v2;
++++++++c->v3+=+*v3;
++++++++return;
++++++++}
++++det+=+1/det;
++++c->center.xw+=+(bc*(cy-dy)-cd*(by-cy))*det;
++++c->center.yw+=+((bx-cx)*cd-(cx-dx)*bc)*det;
++++cx+=+c->center.xw;+cy+=+c->center.yw;
++++c->radius+=+sqrt((cx-bx)*(cx-bx)%2B(cy-by)*(cy-by));
}|3535556|entityMap|0|LINK|mutability|MUTABLE|url|http://mathforum.org/library/drmath/view/54323.html^0|J|O|0|0|3|J|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

Here's my Java port, dodging the error condition when the determinant disappears with a very elegant <code>IllegalArgumentException</code>, my approach to coping with the "points are two far apart" or "points lie on a line" conditions. Also, this computes the radius (and copes with exceptional conditions) which your intersecting-slopes approach will not do.

<pre><code>public class CircleThree
{ 
 static final double TOL = 0.0000001;

 public static Circle circleFromPoints(final Point p1, final Point p2, final Point p3)
 {
 final double offset = Math.pow(p2.x,2) + Math.pow(p2.y,2);
 final double bc = ( Math.pow(p1.x,2) + Math.pow(p1.y,2) - offset )/2.0;
 final double cd = (offset - Math.pow(p3.x, 2) - Math.pow(p3.y, 2))/2.0;
 final double det = (p1.x - p2.x) * (p2.y - p3.y) - (p2.x - p3.x)* (p1.y - p2.y); 

 if (Math.abs(det) &lt; TOL) { throw new IllegalArgumentException("Yeah, lazy."); }

 final double idet = 1/det;

 final double centerx = (bc * (p2.y - p3.y) - cd * (p1.y - p2.y)) * idet;
 final double centery = (cd * (p1.x - p2.x) - bc * (p2.x - p3.x)) * idet;
 final double radius = 
 Math.sqrt( Math.pow(p2.x - centerx,2) + Math.pow(p2.y-centery,2));

 return new Circle(new Point(centerx,centery),radius);
 }

 static class Circle
 {
 final Point center;
 final double radius;
 public Circle(Point center, double radius)
 {
 this.center = center; this.radius = radius;
 }
 @Override 
 public String toString()
 {
 return new StringBuilder().append("Center= ").append(center).append(", r=").append(radius).toString();
 }
 }

 static class Point
 {
 final double x,y;

 public Point(double x, double y)
 {
 this.x = x; this.y = y;
 }
 @Override
 public String toString()
 {
 return "("+x+","+y+")";
 }

 }

 public static void main(String[] args)
 {
 Point p1 = new Point(0.0,1.0);
 Point p2 = new Point(1.0,0.0);
 Point p3 = new Point(2.0,1.0);
 Circle c = circleFromPoints(p1, p2, p3);
 System.out.println(c);
 }

}
</code></pre>

See <a href="http://mathforum.org/library/drmath/view/54323.html" rel="noreferrer">algorithm from here</a>:

<pre><code>void circle_vvv(circle *c)
{
 c-&gt;center.w = 1.0;
 vertex *v1 = (vertex *)c-&gt;c.p1;
 vertex *v2 = (vertex *)c-&gt;c.p2;
 vertex *v3 = (vertex *)c-&gt;c.p3;
 float bx = v1-&gt;xw; float by = v1-&gt;yw;
 float cx = v2-&gt;xw; float cy = v2-&gt;yw;
 float dx = v3-&gt;xw; float dy = v3-&gt;yw;
 float temp = cx*cx+cy*cy;
 float bc = (bx*bx + by*by - temp)/2.0;
 float cd = (temp - dx*dx - dy*dy)/2.0;
 float det = (bx-cx)*(cy-dy)-(cx-dx)*(by-cy);
 if (fabs(det) &lt; 1.0e-6) {
 c-&gt;center.xw = c-&gt;center.yw = 1.0;
 c-&gt;center.w = 0.0;
 c-&gt;v1 = *v1;
 c-&gt;v2 = *v2;
 c-&gt;v3 = *v3;
 return;
 }
 det = 1/det;
 c-&gt;center.xw = (bc*(cy-dy)-cd*(by-cy))*det;
 c-&gt;center.yw = ((bx-cx)*cd-(cx-dx)*bc)*det;
 cx = c-&gt;center.xw; cy = c-&gt;center.yw;
 c-&gt;radius = sqrt((cx-bx)*(cx-bx)+(cy-by)*(cy-by));
}
</code></pre>

blocks|key|4946950|text|当我在这个问题上徘徊时，我正在寻找一个类似的算法。使用您的代码，但发现这在斜率为0或无穷大的情况下不起作用(当xDelta_a或xDelta_b为0时可能为真)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4946951|我修正了算法，下面是我的代码。注意:我使用的是objective-c编程语言，并且正在修改点值初始化的代码，所以如果这是错误的，我相信在java中工作的程序员可以纠正它。然而，逻辑对所有人都是一样的(上帝保佑算法！！：)|4946952|就我自己的功能测试而言，它工作得非常好。如果逻辑有误，请随时告诉我。|4946953|pt+circleCenter(pt+A,+pt+B,+pt+C)+{

float+yDelta_a+=+B.y+-+A.y;
float+xDelta_a+=+B.x+-+A.x;
float+yDelta_b+=+C.y+-+B.y;
float+xDelta_b+=+C.x+-+B.x;
pt+center+=+P(0,0);

float+aSlope+=+yDelta_a/xDelta_a;
float+bSlope+=+yDelta_b/xDelta_b;

pt+AB_Mid+=+P((A.x%2BB.x)/2,+(A.y%2BB.y)/2);
pt+BC_Mid+=+P((B.x%2BC.x)/2,+(B.y%2BC.y)/2);

if(yDelta_a+==+0)+++++++++//aSlope+==+0
{
++++center.x+=+AB_Mid.x;
++++if+(xDelta_b+==+0)+++++++++//bSlope+==+INFINITY
++++{
++++++++center.y+=+BC_Mid.y;
++++}
++++else
++++{
++++++++center.y+=+BC_Mid.y+%2B+(BC_Mid.x-center.x)/bSlope;
++++}
}
else+if+(yDelta_b+==+0)+++++++++++++++//bSlope+==+0
{
++++center.x+=+BC_Mid.x;
++++if+(xDelta_a+==+0)+++++++++++++//aSlope+==+INFINITY
++++{
++++++++center.y+=+AB_Mid.y;
++++}
++++else
++++{
++++++++center.y+=+AB_Mid.y+%2B+(AB_Mid.x-center.x)/aSlope;
++++}
}
else+if+(xDelta_a+==+0)++++++++//aSlope+==+INFINITY
{
++++center.y+=+AB_Mid.y;
++++center.x+=+bSlope*(BC_Mid.y-center.y)+%2B+BC_Mid.x;
}
else+if+(xDelta_b+==+0)++++++++//bSlope+==+INFINITY
{
++++center.y+=+BC_Mid.y;
++++center.x+=+aSlope*(AB_Mid.y-center.y)+%2B+AB_Mid.x;
}
else
{
++++center.x+=+(aSlope*bSlope*(AB_Mid.y-BC_Mid.y)+-+aSlope*BC_Mid.x+%2B+bSlope*AB_Mid.x)/(bSlope-aSlope);
++++center.y+=+AB_Mid.y+-+(center.x+-+AB_Mid.x)/aSlope;
}

return+center;
}|code-block|syntax|javascript|4946954|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I was looking for a similar algorithm when I hovered over this question.
Took your code but found that this will not work in cases when where either of the slope is 0 or infinity (can be true when xDelta_a or xDelta_b is 0).

I corrected the algorithm and here is my code.
Note: I used objective-c programming language and am just changing the code for point value initialization, so if that is wrong, I am sure programmer working in java can correct it. The logic, however, is the same for all (God bless algorithms!! :))

Works perfectly fine as far as my own functional testing is concerned.
Please let me know if logic is wrong at any point.

<pre><code>pt circleCenter(pt A, pt B, pt C) {

float yDelta_a = B.y - A.y;
float xDelta_a = B.x - A.x;
float yDelta_b = C.y - B.y;
float xDelta_b = C.x - B.x;
pt center = P(0,0);

float aSlope = yDelta_a/xDelta_a;
float bSlope = yDelta_b/xDelta_b;

pt AB_Mid = P((A.x+B.x)/2, (A.y+B.y)/2);
pt BC_Mid = P((B.x+C.x)/2, (B.y+C.y)/2);

if(yDelta_a == 0) //aSlope == 0
{
 center.x = AB_Mid.x;
 if (xDelta_b == 0) //bSlope == INFINITY
 {
 center.y = BC_Mid.y;
 }
 else
 {
 center.y = BC_Mid.y + (BC_Mid.x-center.x)/bSlope;
 }
}
else if (yDelta_b == 0) //bSlope == 0
{
 center.x = BC_Mid.x;
 if (xDelta_a == 0) //aSlope == INFINITY
 {
 center.y = AB_Mid.y;
 }
 else
 {
 center.y = AB_Mid.y + (AB_Mid.x-center.x)/aSlope;
 }
}
else if (xDelta_a == 0) //aSlope == INFINITY
{
 center.y = AB_Mid.y;
 center.x = bSlope*(BC_Mid.y-center.y) + BC_Mid.x;
}
else if (xDelta_b == 0) //bSlope == INFINITY
{
 center.y = BC_Mid.y;
 center.x = aSlope*(AB_Mid.y-center.y) + AB_Mid.x;
}
else
{
 center.x = (aSlope*bSlope*(AB_Mid.y-BC_Mid.y) - aSlope*BC_Mid.x + bSlope*AB_Mid.x)/(bSlope-aSlope);
 center.y = AB_Mid.y - (center.x - AB_Mid.x)/aSlope;
}

return center;
}
</code></pre>

blocks|key|3535628|text|public+Vector2+CarculateCircleCenter(Vector2+p1,+Vector2+p2,+Vector2+p3)
{
++++if+(
++++++++p2.x+-+p1.x+==+0+%7C%7C
++++++++p3.x+-+p2.x+==+0+%7C%7C
++++++++p2.y+-+p1.y+==+0+%7C%7C
++++++++p3.y+-+p2.y+==+0
++++)+return+null;

++++Vector2+center+=+new+Vector2();
++++float+ma+=+(p2.y+-+p1.y)+/+(p2.x+-+p1.x);
++++float+mb+=+(p3.y+-+p2.y)+/+(p3.x+-+p2.x);
++++center.x+=+(ma+*+mb+*+(p1.y+-+p3.y)+%2B+mb+*+(p1.x+-+p2.x)+-+ma+*+(p2.x+%2B+p3.x))+/+(2+*+(mb+-+ma));
++++center.y+=+(-1+/+ma)+*+(center.x+-+(p1.x+%2B+p2.x)+*+0.5)+%2B+(p1.y+%2B+p2.y)+*+0.5;
++++return+center;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3535629|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public Vector2 CarculateCircleCenter(Vector2 p1, Vector2 p2, Vector2 p3)
{
 if (
 p2.x - p1.x == 0 ||
 p3.x - p2.x == 0 ||
 p2.y - p1.y == 0 ||
 p3.y - p2.y == 0
 ) return null;

 Vector2 center = new Vector2();
 float ma = (p2.y - p1.y) / (p2.x - p1.x);
 float mb = (p3.y - p2.y) / (p3.x - p2.x);
 center.x = (ma * mb * (p1.y - p3.y) + mb * (p1.x - p2.x) - ma * (p2.x + p3.x)) / (2 * (mb - ma));
 center.y = (-1 / ma) * (center.x - (p1.x + p2.x) * 0.5) + (p1.y + p2.y) * 0.5;
 return center;
}
</code></pre>

blocks|key|3535678|text|很抱歉，我的回答来晚了。当两个点形成一条垂直线时，任何使用“斜率”的解决方案都将失败，因为斜率将是无穷大的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3535679|以下是2019年的一个简单而健壮的解决方案，它始终正常工作：|3535680|public+static+boolean+circleCenter(double[]+p1,+double[]+p2,+double[]+p3,+double[]+center)+{
++++double+ax+=+(p1[0]+%2B+p2[0])+/+2;
++++double+ay+=+(p1[1]+%2B+p2[1])+/+2;
++++double+ux+=+(p1[1]+-+p2[1]);
++++double+uy+=+(p2[0]+-+p1[0]);
++++double+bx+=+(p2[0]+%2B+p3[0])+/+2;
++++double+by+=+(p2[1]+%2B+p3[1])+/+2;
++++double+vx+=+(p2[1]+-+p3[1]);
++++double+vy+=+(p3[0]+-+p2[0]);
++++double+dx+=+ax+-+bx;
++++double+dy+=+ay+-+by;
++++double+vu+=+vx+*+uy+-+vy+*+ux;
++++if+(vu+==+0)
++++++++return+false;+//+Points+are+collinear,+so+no+unique+solution
++++double+g+=+(dx+*+uy+-+dy+*+ux)+/+vu;
++++center[0]+=+bx+%2B+g+*+vx;
++++center[1]+=+by+%2B+g+*+vy;
++++return+true;
}|code-block|syntax|javascript|3535681|当且仅当这3个点共线时，上述代码才会返回"false“。|3535682|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I am sorry my answer is late. Any solution using "slope" will fail when two of the points form a vertical line, because the slope will be infinite.

Here is a simple robust solution for 2019 that always works correctly:

<pre><code>public static boolean circleCenter(double[] p1, double[] p2, double[] p3, double[] center) {
 double ax = (p1[0] + p2[0]) / 2;
 double ay = (p1[1] + p2[1]) / 2;
 double ux = (p1[1] - p2[1]);
 double uy = (p2[0] - p1[0]);
 double bx = (p2[0] + p3[0]) / 2;
 double by = (p2[1] + p3[1]) / 2;
 double vx = (p2[1] - p3[1]);
 double vy = (p3[0] - p2[0]);
 double dx = ax - bx;
 double dy = ay - by;
 double vu = vx * uy - vy * ux;
 if (vu == 0)
 return false; // Points are collinear, so no unique solution
 double g = (dx * uy - dy * ux) / vu;
 center[0] = bx + g * vx;
 center[1] = by + g * vy;
 return true;
}
</code></pre>

The above code will return "false" if and only if the 3 points are collinear.

blocks|key|1287093|text|def+circle_mid_point(list_b):
list_k+=+[]
for+i+in+range(3):
++++for+j+in+range(1,4):
++++++++if+i%2Bj+<=3:
++++++++++++midpoint_x1+=++(list_b[i][0]+%2B+list_b[i%2Bj][0])/2
++++++++++++midpoint_y1+=+(list_b[i][1]+%2B+list_b[i+%2B+j][1])+/+2
++++++++++++list_k.append([midpoint_x1,midpoint_y1])+#+list+of+all+the+midpoints+of+the+lines

for+k+in+range(len(list_k)):
++++for+j+in+range(1,len(list_k)-1):
++++++++if+list_k[k]+==+list_k[k%2Bj]:+#at+centre+only+two+midpoints+will+have+the+same+value
++++++++++++return+list_k[k]

k+=+circle_mid_point([[0,1],[1,0],[-1,0],[0,-1]])
print(k)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1287094|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def circle_mid_point(list_b):
list_k = []
for i in range(3):
 for j in range(1,4):
 if i+j &lt;=3:
 midpoint_x1 = (list_b[i][0] + list_b[i+j][0])/2
 midpoint_y1 = (list_b[i][1] + list_b[i + j][1]) / 2
 list_k.append([midpoint_x1,midpoint_y1]) # list of all the midpoints of the lines

for k in range(len(list_k)):
 for j in range(1,len(list_k)-1):
 if list_k[k] == list_k[k+j]: #at centre only two midpoints will have the same value
 return list_k[k]

k = circle_mid_point([[0,1],[1,0],[-1,0],[0,-1]])
print(k)
</code></pre>

I have three points on the circumference of a circle:

<pre><code>pt A = (A.x, A.y);
pt B = (B.x, B.y);
pt C = (C.x, C.y);
</code></pre>

How do I calculate the center of the circle?

Implementing it in Processing (Java).

I found the answer and implemented a working solution:

<pre><code> pt circleCenter(pt A, pt B, pt C) {

 float yDelta_a = B.y - A.y;
 float xDelta_a = B.x - A.x;
 float yDelta_b = C.y - B.y;
 float xDelta_b = C.x - B.x;
 pt center = P(0,0);

 float aSlope = yDelta_a/xDelta_a;
 float bSlope = yDelta_b/xDelta_b; 
 center.x = (aSlope*bSlope*(A.y - C.y) + bSlope*(A.x + B.x)
 - aSlope*(B.x+C.x) )/(2* (bSlope-aSlope) );
 center.y = -1*(center.x - (A.x+B.x)/2)/aSlope + (A.y+B.y)/2;

 return center;
 }
</code></pre>

What is the algorithm for finding the center of a circle from three points?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我在圆的圆周上有三个点：pt A = (A.x, A.y);pt B = (B.x, B.y);pt C = (C.x, C.y);如何计算圆的圆心？在处理中实现它(Java)。我找到了答案并实现了一个可行的解决方案： pt circleCenter(pt A, pt B, pt C) {    float yDelt...

问从三个点找到圆的圆心的算法是什么？
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问从三个点找到圆的圆心的算法是什么？EN